Mastering Physics Solutions Chapter 7 Work And Kinetic Energy
Chapter 7 Work And Kinetic Energy Q.1CQ
Is it possible to do work on an object that remains at rest?
Solution:
No.
We know that work is said to be done only when a body moves a certain distance in the direction of an applied force. In other words, if no external force is applied or the body fails to move in the direction of an applied force, the work done is said to be zero.
Consider a block of mass m (kg) moving a distance d (m) by application of force F (N) in the direction of the motion. Then, work done
W = F·d = force distance
If d = 0 [i.e., the body is at rest]
W = F 0
= 0
Therefore, it is not possible to do work on an object that is at rest.
Chapter 7 Work And Kinetic Energy Q.1P
The International Space Station orbits the Earth in an approximately circular orbit at a height of h = 375 km above the Earth’s surface. In one complete orbit, is the work done by the Earth on the space station positive, negative, or zero? Explain.
Solution:
The work done by Earth on the space station is zero. This is because during the movement of the satellite, the force acting on it is always perpendicular to the direction of motion. If the force does not have non-zero components along the direction of motion, the work done is zero.
Chapter 7 Work And Kinetic Energy Q.2CQ
A friend makes the statement, “Only the total force acting on an object can do work.” Is this statement true or false? If it is true, state why; if it is false, give a counterexample.
Solution:
Chapter 7 Work And Kinetic Energy Q.2P
A pendulum bob swings from point I to point II along the circular arc indicated in Figure. (a) Is the work done on the bob by gravity positive, negative, or zero? Explain. (b) Is the work done on the bob by the string positive, negative, or zero? Explain.
Solution:
(a) When the pendulum bob swings from point I to point II along the circular arc, its displacement is downward. This is in the direction of force of gravity. Therefore the work done by the gravity is positive when the pendulum bob swings from point I to pint II along the circular arc.
(b) When the pendulum bob swings from point I to point II along the circular arc, the tension in the string is always perpendicular to its direction of motion. Therefore the work done by the string is zero when the pendulum bob swings from point I to pint II along the circular arc.
Chapter 7 Work And Kinetic Energy Q.3CQ
A friend makes the statement, “A force that is always perpendicular to the velocity of a particle does no work on the particle.” Is this statement true or false? If it is hue, state why; if it is false, give a counterexample.
Solution:
True. A force that is always perpendicular to displacement does not have a non-zero component along the direction of motion. As a result, work will not be done on the particle.
Work done W=(F cosθ)d
W is positive if F has a component in the direction of motion.
W is positive if the angle between the force F and displacement d is -90< <90.
Chapter 7 Work And Kinetic Energy Q.3P
A pendulum bob swings from point II to point III along the circular arc indicated in Figure. (a) Is the work done on the bob by gravity positive, negative, or zero? Explain. (b) Is the work done on the bob by the string positive, negative, or zero? Explain.
Solution:
(a) When the pendulum bob swings from point II to point III along the circular arc, its displacement is upward. This is opposite to the direction of force of gravity. Therefore the work done by the gravity is negative when the pendulum bob swings from point II to pint III along the circular arc.
(b) When the pendulum bob swings from point II to point III along the circular arc, the tension in the string is always perpendicular to its direction of motion. Therefore the work done by the string is zero when the pendulum bob swings from point II to pint III along the circular arc.
Chapter 7 Work And Kinetic Energy Q.4CQ
The net work done on a certain object is zero. What can you say about its speed?
Solution:
The work done is equal to the change in kinetic energy. So if the net work done on an object is zero, its change in kinetic energy is also zero. Thus, the speed of the object will remain the same.
Chapter 7 Work And Kinetic Energy Q.4P
A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how much work has she done?
Solution:
Chapter 7 Work And Kinetic Energy Q.5CQ
To get out of bed in the morning, do you have to do work? Explain.
Solution:
Yes, we have to do work against the force of gravity to get out of bed in the morning because the constant force of gravity acts downward on our body when we are sleeping in bed. To get up we have to apply force upward; thus, we are doing work against the force of gravity.
Chapter 7 Work And Kinetic Energy Q.5P
Children in a tree house lift a small dog in a basket 4.70 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket?
Solution:
Chapter 7 Work And Kinetic Energy Q.6CQ
Give an example of a frictional force doing negative work.
Solution:
Frictional forces do negative work whenever they act in a direction that opposes the motion.
For example, friction does negative work when you push a box across the floor, or when you hit the brakes while driving.
Chapter 7 Work And Kinetic Energy Q.6P
Early one October, you go to a pumpkin patch to select your Halloween pumpkin. You lift the 3.2-kg pumpkin to a height of 1.2 in, then carry it 50.0 m (on level ground) to the check-out stand. (a) Calculate the work you do on the pumpkin as you lift it from the ground. (b) How much work do you do on the pumpkin as you carry it from the field?
Solution:
Chapter 7 Work And Kinetic Energy Q.7CQ
Give an example of a frictional force doing positive work.
Solution:
The force of friction always opposes relative motion, but if the body moves along the direction of applied force, for ◊ = 0º, the work done is positive.
So when man walks on the ground, the frictional force between his shoes and the ground does positive work whenever he begins to walk, as ◊ = 0º.
Chapter 7 Work And Kinetic Energy Q.7P
The coefficient of kinetic friction between a suitcase and the floor is 0.272. If the suitcase has a mass of 71.5 kg, how far can it be pushed across the level floor with 642 J of work?
Solution:
Chapter 7 Work And Kinetic Energy Q.8CQ
A ski boat moves with constant velocity. Is the net force acting on the boat doing work? Explain.
Solution:
Since the velocity of the boat is constant, there will not be any change in the kinetic energy of the boat.
Thus, the change in kinetic energy = 0 joules ————- (1)
Work done = change in kinetic energy —————- (2)
From equations (1) and (2),
work done = 0 joules ——————– (3)
Work done = net force displacement ————— (4)
Since the boat is moving at a constant speed, then from equations (3) and (4),
net force = 0 newtons
Thus, the net force is not doing any work on the boat.
Chapter 7 Work And Kinetic Energy Q.8P
You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) How much work do you do on the can of paint? (b) You hold the can stationary for half a minute, waiting for a friend on a ladder to take it. How much work do you do during this time? (c) Your friend decides against the paint, so you lower it back to the ground. How much work do you do on the can as you lower it?
Solution:
Mass of the can m = 3.4 kg
Chapter 7 Work And Kinetic Energy Q.9CQ
A package rests on the floor of an elevator that is rising with constant speed. The elevator exerts an upward normal force on the package, and hence does positive work on it. Why doesn’t the kinetic energy of the package increase?
Solution:
Chapter 7 Work And Kinetic Energy Q.9P
A tow rope, parallel to the water, pulls a water skier directly behind the boat with constant velocity for a distance of 65 m before the skier falls. The tension in the rope is 120 N. (a) Is the work done on the skier by the rope positive, negative, or zero? Explain. (b) Calculate the work done by the rope on the skier.
Solution:
Chapter 7 Work And Kinetic Energy Q.10CQ
An object moves with constant velocity. Is it safe to conclude that no force acts on the object? Why, or why not?
Solution:
Chapter 7 Work And Kinetic Energy Q.10P
In the situation described in the previous problem, (a) is the work done on the boat by the rope positive, negative, or zero? Explain. (b) Calculate the work done by the rope on the boat.
Solution:
Chapter 7 Work And Kinetic Energy Q.11CQ
Engine 1 does twice the work of engine 2. Is it correct to conclude that engine 1 produces twice as much power as engine 2? Explain.
Solution:
The power produced is defined as the rate of the work done. Here it is mentioned that engine 1 does twice the amount of work as engine 2 but their individual time of doing that work is not specified.
Hence, it is incorrect to conclude that engine 1 produces twice the amount of power as produced by engine 2.
Chapter 7 Work And Kinetic Energy Q.11P
A child pulls a friend in a little red wagon with constant speed. If the child pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 25° above the horizontal, how much work does the child do on the wagon?
Solution:
Chapter 7 Work And Kinetic Energy Q.12CQ
Engine 1 produces twice the power of engine 2. Is it correct to conclude that engine 1 does twice as much work as engine 2? Explain.
Solution:
No.
Power depends both on the amount of work done by the engine, and the amount of time during which the work is performed.
Chapter 7 Work And Kinetic Energy Q.12P
A 51-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 43.5° above the horizontal. If the tension in the rope is 115 N, how much work is done on the crate to move it 8.0 m?
Solution:
Chapter 7 Work And Kinetic Energy Q.13P
To clean a floor, a janitor pushes on a mop handle with a force of 50.0 N. (a) If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop 0.50 m? (b) If the angle the mop handle makes with the horizontal is increased to 65°, does the work done by the janitor increase, decrease, or stay the same? Explain.
Solution:
Chapter 7 Work And Kinetic Energy Q.14P
A small plane tows a glider at constant speed and altitude. If the plane does 2.00 × 105 J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the tow rope and the horizontal?
Solution:
Chapter 7 Work And Kinetic Energy Q.15P
Solution:
Chapter 7 Work And Kinetic Energy Q.16P
Solution:
Chapter 7 Work And Kinetic Energy Q.17P
Water skiers often ride to one side of the center line of a boat, as shown in Figure. In this case, the ski boat is traveling at 15 m/s and the tension in the rope is 75 N. If the boat does 3500 J of work on the skier in 50.0 m, what is the angle θ between the tow rope and the center line of the boat?
Solution:
Chapter 7 Work And Kinetic Energy Q.18P
A pitcher throws a ball at 90 mi /h and the catcher stops it in her glove. (a) Is the work done on the ball by the pitcher positive, negative, or zero? Explain. (b) Is the work done on the ball by the catcher positive, negative, or zero? Explain.
Solution:
a) The pitcher does positive work on the ball because the direction of force is along the direction of displacement of the ball.
b) The catcher does negative work on the ball by exerting a force in the direction opposite to the motion of the ball, in order to stop the ball.
Chapter 7 Work And Kinetic Energy Q.19P
How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution:
Chapter 7 Work And Kinetic Energy Q.20P
Skylab’s Reentry When Skylab reentered the Earth’s atmosphere on July 11, 1979, it broke into a myriad of pieces. One of the largest fragments was a I770-kg lead-lined film vault, and it landed with an estimated speed of 120 m/s. What was the kinetic energy of the film vault when it landed?
Solution:
Chapter 7 Work And Kinetic Energy Q.21P
A 9.50-g bullet has a speed of 1.30 km/s. (a) What is its kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled?
Solution:
Chapter 7 Work And Kinetic Energy Q.22P
The work W0 accelerates a car from 0 to 50 km/h. (a) Is the work required to accelerate the car from 50 km/h to 150 km/h equal to 2W0, 3W0, 8W0, or 91w0? (b) Choose the best explanation from among the following:
I. The work to accelerate the car depends on the speed squared.
II. The final speed is three times the speed that was produced by the work W0.
III. The increase in speed from 50 km/h to 150 km/h is twice the increase in speed from 0 to 50 km/h.
Solution:
Chapter 7 Work And Kinetic Energy Q.23P
Jogger A has a mass m and a speed v, jogger B has a mass m/2 and a speed 3v, jogger C has a mass 3m and a speed v/2, and jogger D has a mass 4m and a speed v/2. Rank the joggers in order of increasing kinetic energy. Indicate ties where appropriate.
Solution:
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Chapter 7 Work And Kinetic Energy Q.24P
A 0.14-kg pinecone falls 16 m to the ground, where it lands with a speed of 13 m/s. (a) With what speed would the pinecone have landed if there had been no air resistance? (b) Did air resistance do positive work, negative work, or zero work on the pinecone? Explain.
Solution:
Chapter 7 Work And Kinetic Energy Q.25P
In the previous problem, (a) how much work was done on the pinecone by air resistance? (b) What was the average force of air resistance exerted on the pinecone?
Solution:
The negative sign indicates that the force acts in the upward direction.
Chapter 7 Work And Kinetic Energy Q.26P
At t = 1.0 s, a 0.40-kg object is falling with a speed of 6.0 m/s. At t = 2.0 s, it has a kinetic energy of 25 J. (a) What is the kinetic energy of the object at t = 1.0 s? (b) What is the speed of the object at t = 2.0 s? (c) How much work was done on the object between t = 1.0 s and t = 2.0 s?
Solution:
Chapter 7 Work And Kinetic Energy Q.27P
After hitting a long fly ball that goes over the right fielder’s head and lands in the outfield, the batter decides to keep going past second base and try for third base. The 62.0-kg player begins sliding 3.40 m from the base with a speed of 4.35 m/s. If the player comes to rest at third base, (a) how much work was done on the player by friction? (b) What was the coefficient of kinetic friction between the player and the ground?
Solution:
Chapter 7 Work And Kinetic Energy Q.28P
A 1100-kg car coasts on a horizontal road with a speed of 19 m/s. After crossing an unpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net work done on the car positive, negative, or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section.
Solution:
Chapter 7 Work And Kinetic Energy Q.29P
(a) In the previous problem, the car’s speed decreased by 7.0 m/s as it coasted across a sandy section of road 32 m long. If the sandy portion of the road had been only 16 m long, would the car’s speed have decreased by 3.5 m/s, more than 3.5 m/s, or less than 3.5 m/s? Explain. (b) Calculate the change in speed in this case.
Solution:
Chapter 7 Work And Kinetic Energy Q.30P
A 65-kg bicyclist rides his 8.8-kg bicycle with a speed of 14 m/s. (a) How much work must be done by the brakes to bring the bike and rider to a stop? (b) How far does the bicycle travel if it takes 4.0 s to come to rest? (c) What is the magnitude of the braking force?
Solution:
Chapter 7 Work And Kinetic Energy Q.31P
A block of mass m and speed v collides with a spring, compressing it a distance △x. What is the compression of the spring if the force constant of the spring is increased by a factor of four?
Solution:
Chapter 7 Work And Kinetic Energy Q.32P
A spring with a force constant of 3.5 × 104 N/m is initially at its equilibrium length. (a) How much work must you do to stretch the spring 0.050 m? (b) How much work must you do to compress it 0.050 m?
Solution:
Chapter 7 Work And Kinetic Energy Q.33P
A 1.2-kg block is held against a spring of force constant 1.0 × 104 N/m, compressing it a distance of 0.15 m. How fast is the block moving after it is released and the spring pushes it away?
Solution:
Chapter 7 Work And Kinetic Energy Q.34P
Initially sliding with a speed of 2.2 m/s, a 1.8-kg block collides with a spring and compresses it 0.31 m before coming to rest. What is the force constant of the spring?
Solution:
Chapter 7 Work And Kinetic Energy Q.35P
Solution:
Chapter 7 Work And Kinetic Energy Q.36P
Solution:
Chapter 7 Work And Kinetic Energy Q.37P
CE A block of mass m and speed v collides with a spring, compressing it a distance △x. What is the compression of the spring if the mass of the block is halved and its speed is doubled?
Solution:
Chapter 7 Work And Kinetic Energy Q.38P
To compress spring 1 by 0.20 m takes 150 J of work. Stretching spring 2 by 0.30 m requires 210 J of work. Which spring is stiffer?
Solution:
38ps
Work done to compress spring 1 by x = 0.2 m is W = 150 J We know that
Chapter 7 Work And Kinetic Energy Q.39P
It takes 180 J of work to compress a certain spring 0.15 m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.15 m, does it take 180 J, more than 180 J, or less than 180 J? Verify your answer with a calculation.
Solution:
Chapter 7 Work And Kinetic Energy Q.40P
Solution:
Chapter 7 Work And Kinetic Energy Q.41P
A block is acted on by a force that varies as (2.0 × 104 N/m)x for 0 ≤ x ≤ 0.21 m, and then remains constant at 4200 N for larger x. How much work does the force do on the block in moving it (a) from x = 0 to x = 0.30 m, or (b) from x = 0.10 m to x = 0.40 m?
Solution:
Chapter 7 Work And Kinetic Energy Q.42P
CE Force F1 does 5 J of work in 10 seconds, force F2 does 3 J of work in 5 seconds, force F3 does 6 J of work in 18 seconds, and force F4 does 25 J of work in 125 seconds. Rank these forces in order of increasing power they produce. Indicate ties where appropriate.
Solution:
Chapter 7 Work And Kinetic Energy Q.43P
Climbing the Empire State Building A new record for running the stairs of the Empire State Building was set on February 3, 2003. The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. If the height gain of the step was 0.20 m, and the mass of the runner was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.
Solution:
Chapter 7 Work And Kinetic Energy Q.44P
How many joules of energy are in a kilowatt-hour?
Solution:
Chapter 7 Work And Kinetic Energy Q.45P
Calculate the power output of a 1.4-g fly as it walks straight tip a windowpane at 2.3 cm/s.
Solution:
Chapter 7 Work And Kinetic Energy Q.46P
An ice cube is placed in a microwave oven. Suppose the oven delivers 105 W of power to the ice cube and that it takes 32,200 J to melt it. How long docs it take for the ice cube to melt?
Solution:
Chapter 7 Work And Kinetic Energy Q.47P
You raise a bucket of water from the bottom of a deep well. If your power output is 108 W, and the mass of the bucket and the water in it is 5.00 kg, with what speed can you raise the bucket? Ignore the weight of the rope.
Solution:
Chapter 7 Work And Kinetic Energy Q.48P
In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water the second from below deck up a height of 2.00 m and over the side. What is the minimum horse power motor that can be used to save the ship?
Solution:
Chapter 7 Work And Kinetic Energy Q.49P
A kayaker paddles with a power output of 50.0 W to maintain a steady speed of 1.50 m/s. (a) Calculate the resistive force exerted by the water on the kayak. (b) If the kayaker doubles her power output, and the resistive force due to the water remains the same, by what factor does the kayaker’s speed change?
Solution:
Chapter 7 Work And Kinetic Energy Q.50P
Human-Powered Flight Human-powered aircraft re quire a pilot to pedal, as in a bicycle, and produce a sustained power output of about 0.30 hp. The Gossamer Albatross flew across the English Channel on June 12, 1979, in 2 h 49 min. (a) How much energy did the pilot expend during the flight? (b) How many Snickers candy bars (280 Cal per bar) would the pilot have to consume to be “fueled up” for the flight? [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.]
Solution:
Chapter 7 Work And Kinetic Energy Q.51P
A grandfather clock is powered by the descent of a 4.35-kg weight. (a) If the weight descends through a distance of 0.760 m in 3.25 days, how much power docs it deliver to the clock? (b) To increase the power delivered to the clock, should the time it takes for the mass to descend be increased or decreased? Explain.
Solution:
Chapter 7 Work And Kinetic Energy Q.52P
The Power You Produce Estimate the power you produce in running up a flight of stairs. Give your answer in horsepower.
Solution:
Chapter 7 Work And Kinetic Energy Q.53P
A certain car can accelerate from rest to the speed v in T seconds. If the power output of the car remains constant, (a) how long does it take for the car to accelerate from v to 2v? (b) How fast is the car moving at 2T seconds after starting?
Solution:
Chapter 7 Work And Kinetic Energy Q.54GP
Solution:
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Chapter 7 Work And Kinetic Energy Q.55GP
A youngster rides on a skateboard with a speed of 2 m/s. After a force acts on the youngster, her speed is 3 m/s. Was the work done by the force positive, negative, or zero? Explain.
Solution:
Chapter 7 Work And Kinetic Energy Q.56GP
Predict/Explain A car is accelerated by a constant force, F. The distance required to accelerate the car from rest to the speed v is △x. (a) Is the distance required to accelerate the car from the speed v to the speed 2v equal to △x, 2△x, 3△x, or 4△x? (b) Choose the best explanation from among the following:
I. The final speed is twice the initial speed.
II. The increase in speed is the same in each case.
III. Work is force times distance, and work depends on the speed squared.
Solution:
Chapter 7 Work And Kinetic Energy Q.57GP
Car 1 has four times the mass of car 2, but they both have the same kinetic energy. If the speed of car 2 is v, is the speed of car 1 equal to v/4, v/2, 2v, or 4v? Explain.
Solution:
Chapter 7 Work And Kinetic Energy Q.58GP
Muscle Cells Biological muscle cells can be thought of as nanomotors that use the chemical energy of ATP to produce mechanical work. Measurements show that the active proteins within a muscle cell (such as myosin and actin) can produce a force of about 7.5 pN and displacements of 8.0 run. How much work is done by such proteins?
Solution:
Chapter 7 Work And Kinetic Energy Q.59GP
When you take a bite out of an apple, you do about 19 J of work. Estimate (a) the force and (b) the power produced by your jaw muscles during the bite.
Solution:
Chapter 7 Work And Kinetic Energy Q.60GP
A Mountain bar has a mass of 0.045 kg and a calorie rating of 210 Cal. What speed would this candy bar have if its kinetic energy were equal to its metabolic energy? [See the note following Problem.]
Human-Powered Flight Human-powered aircraft re quire a pilot to pedal, as in a bicycle, and produce a sustained power output of about 0.30 hp. The Gossamer Albatross flew across the English Channel on June 12, 1979, in 2 h 49 min. (a) How much energy did the pilot expend during the flight? (b) How many Snickers candy bars (280 Cal per bar) would the pilot have to consume to be “fueled up” for the flight? [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.]
Solution:
Chapter 7 Work And Kinetic Energy Q.61GP
A small motor runs a lift that raises a load of bricks weighing 836 N to a height of 10.7 m in 23.2 s. Assuming that the bricks are lifted with constant speed, what is the minimum power the motor must produce?
Solution:
Chapter 7 Work And Kinetic Energy Q.62GP
You push a 67-kg box across a floor where the coefficient of kinetic friction is µk = 0.55. The force you exert is horizontal. (a) How much power is needed to push the box at a speed of 0.50 m/s? (b) How much work do you do if you push the box for 35 s?
Solution:
Chapter 7 Work And Kinetic Energy Q.63GP
The Beating Heart The average power output of the human heart is 1.33 watts. (a) How much energy does the heart produce in a day? (b) Compare the energy found in part (a) with the energy required to walk up a flight of stairs. Estimate the height a person could attain on a set of stairs using nothing more than the daily energy produced by the heart.
Solution:
Chapter 7 Work And Kinetic Energy Q.64GP
The Atmos Clock The Atmos clock (the so-called perpetual motion clock) gets its name from the fact that it runs off pressure variations in the atmosphere, which drive a bellows containing a mixture of gas and liquid ethyl chloride. Because the power to drive these clocks is so limited, they must be very efficient. In fact, a single 60.0-W lightbulb could power 240 million Atmos clocks simultaneously. Find the amount of energy, in joules, required to run an Atmos clock for one day.
Solution:
Chapter 7 Work And Kinetic Energy Q.65GP
The work W0 is required to accelerate a car from rest to the speed v0. How much work is required to accelerate the car (a) from rest to the speed v0/2 and (b) from v0/2 to v0?
Solution:
Chapter 7 Work And Kinetic Energy Q.66GP
A work W0 is required to stretch a certain spring 2 cm from its equilibrium position. (a) How much work is required to stretch the spring 1 cm from equilibrium? (b) Suppose the spring is already stretched 2 cm from equilibrium. How much additional work is required to stretch it to 3 cm from equilibrium?
Solution:
Chapter 7 Work And Kinetic Energy Q.67GP
After a tornado, a 0.55-g straw was found embedded 2.3 cm into the trunk of a tree. If the average force exerted on the straw by the tree was 65 N, what was the speed of the straw when it hit the tree?
Solution:
Chapter 7 Work And Kinetic Energy Q.68GP
You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. What is the kinetic energy of the glove when it is at the height h/2?
Solution:
Chapter 7 Work And Kinetic Energy Q.69GP
Solution:
Chapter 7 Work And Kinetic Energy Q.70GP
Solution:
(C) If the pulling force remains the e, then the work done remains the e. If the increased mass results in a larger friction force and an increase in the pulling force, then the work done increases.
Chapter 7 Work And Kinetic Energy Q.71GP
A 0.19-kg apple falls from a branch 3.5 m above the ground. (a) Does the power delivered to the apple by gravity increase, de crease, or stay the same during the time the apple falls to the ground? Explain. Find the power delivered by gravity to the apple when the apple is (b) 2.5 m and (c) 1.5 m above the ground.
Solution:
Chapter 7 Work And Kinetic Energy Q.72GP
A juggling ball of mass m is thrown straight upward from an initial height h with an initial speed v0. How much work has gravity done on the ball (a) when it reaches its greatest height, h max, and (b) when it reaches ground level? (c) Find an expression for the kinetic energy of the ball as it lands.
Solution:
Chapter 7 Work And Kinetic Energy Q.73GP
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Chapter 7 Work And Kinetic Energy Q.74GP
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Chapter 7 Work And Kinetic Energy Q.75GP
The motor of a ski boat produces a power of 36,600 W to maintain a constant speed of 14.0 m/s. To pull a water skier at the same constant speed, the motor must produce a power of 37,800 W. What is the tension in the rope pulling the skier?
Solution:
Chapter 7 Work And Kinetic Energy Q.76GP
Cookie Power To make a batch of cookies, you mix half a bag of chocolate chips into a bowl of cookie dough, exerting a 21-N force on the stirring spoon. Assume that your force is always in the direction of motion of the spoon. (a) What power is needed to move the spoon at a speed of 0.23 m/s? (b) How much work do you do if you stir the mixture for 1.5 min?
Solution:
Chapter 7 Work And Kinetic Energy Q.77GP
A pitcher accelerates a 0.14-kg hardball from rest to 42.5 m/s in 0.060 s. (a) How much work does the pitcher do on the ball? (b) What is the pitcher’s power output during the pitch? (c) Suppose the ball reaches 42.5 m/s in less than 0.060 s. Is the power produced by the pitcher in this case more than, less than, or the same as the power found in part (b)? Explain.
Solution:
Chapter 7 Work And Kinetic Energy Q.78GP
Catapult Launcher A catapult launcher on an aircraft carrier accelerates a jet from rest to 72 m/s. The work done by the catapult during the launch is 7.6 × 107 J. (a) What is the mass of the jet? (b) If the jet is in contact with the catapult for 2.0 s, what is the power output of the catapult?
Solution:
Chapter 7 Work And Kinetic Energy Q.79GP
Brain Power The human brain consumes about 22 W of power under normal conditions, though more power may be required during exams. (a) How long can one Snickers bar (see the note following Problem) power the normally functioning brain? (b) At what rate must you lift a 3.6-kg container of milk (one gallon) if the power output of your arm is to be 22 W? (c) How long does it take to lift the milk container through a distance of 1.0 m at this rate?
Human-Powered Flight Human-powered aircraft re quire a pilot to pedal, as in a bicycle, and produce a sustained power output of about 0.30 hp. The Gossamer Albatross flew across the English Channel on June 12, 1979, in 2 h 49 min. (a) How much energy did the pilot expend during the flight? (b) How many Snickers candy bars (280 Cal per bar) would the pilot have to consume to be “fueled up” for the flight? [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.]
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Chapter 7 Work And Kinetic Energy Q.80GP
A 1300-kg car delivers a constant 49 hp to the drive wheels. We assume the car is traveling on a level road and that all fractional forces may be ignored. (a) What is the acceleration of this car when its speed is 14 m/s? (b) If the speed of the car is doubled, does its acceleration increase, decrease, or stay the same? Explain. (c) Calculate the car’s acceleration when its speed is 28 m/s.
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Chapter 7 Work And Kinetic Energy Q.81GP
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Chapter 7 Work And Kinetic Energy Q.82GP
Powering a Pigeon A pigeon in flight experiences a force of air resistance given approximately by F = bv2, where v is the flight speed and b is a constant. (a) What are the units of the constant b? (b) What is the largest possible speed of the pigeon if its maximum power output is P? (c) By what factor does the largest possible speed increase if the maximum power is doubled?
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Chapter 7 Work And Kinetic Energy Q.83GP
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Chapter 7 Work And Kinetic Energy Q.84GP
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Chapter 7 Work And Kinetic Energy Q.85GP
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Chapter 7 Work And Kinetic Energy Q.86PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Solution:
From the given figure it is clear that the range of flight speeds for power out put of 9.8W is 7.7m/s to 15m/s
So, option B. is correct.
Chapter 7 Work And Kinetic Energy Q.87PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Solution:
From the figure it is clear that the approximate range of flight speeds would be possible if Microraptor gui could produce 20W of power is 2.5m/s to 25m/s.
This is because below 2.5m/s and above 25m/s, the power out put is greater than 20W.
Therefore option C. is correct.
Chapter 7 Work And Kinetic Energy Q.88PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Solution:
Chapter 7 Work And Kinetic Energy Q.89PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Solution:
Chapter 7 Work And Kinetic Energy Q.90IP
Referring to Figure Suppose the block has a mass of 1.4 kg and an initial speed of 0.62 m/s. (a) What force constant must the spring have if the maximum compression is to be 2.4 cm? (b) If the spring has the force constant found in part (a), find the maximum compression if the mass of the block is doubled and its initial speed is halved.
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Chapter 7 Work And Kinetic Energy Q.91IP
Referring to Figure In the situation shown in Figure (d), a spring with a force constant of 750 N/m is compressed by 4.1 cm. (a) If the speed of the block in Figure (f) is 0.88 m/s, what is its mass? (b) If the mass of the block is doubled, is the final speed greater than, less than, or equal to 0.44 m/s? (c) Find the final speed for the case described in part (b).
Solution:
Chapter 7 Work And Kinetic Energy Q.92IP
Referring to Example Suppose the car has a mass of 1400 kg and delivers 48 hp to the wheels. (a) How long does it take for the car to increase its speed from 15 m/s to 25 m/s? (b) Would the time required to increase the speed from 5.0 m/s to 15 m/s be greater than, less than, or equal to the time found in part (a)? (c) Determine the time required to accelerate from 5.0 m/s to 15 m/s.
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