HW Pract 08 - Work and Energy Key ghjf yguk kg hjjhg hjk hjg hjkg k 578e 7if7i PDF

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AP Physics 1

Name: __________________ Date: ____________ Prd: ___

HW Practice 08 – Work and Energy Key Responses to Questions 1.

“Work” as used in everyday language generally means “energy expended,” which is similar to the way “work” is defined in physics. However, in everyday language, “work” can involve mental or physical energy expended and is not necessarily connected with displacement, as it is in physics. So a student could say she “worked” hard carrying boxes up the stairs to her dorm room (similar in meaning to the physics usage), or that she “worked” hard on a problem set (different in meaning from the physics usage).

2.

No, not if the object is moving in a circle. Work is the product of force and the displacement in the direction of the force. Therefore, a centripetal force, which is perpendicular to the direction of motion, cannot do work on an object moving in a circle.

6.

The kinetic energy increases by a factor of 9, since the kinetic energy is proportional to the square of the speed.

8.

The speed at point C will be less than twice the speed at point B. The force is constant and the displacements are the same, so the same work is done on the block from A to B as from B to C. Since there is no friction, the same work results in the same change in kinetic energy. But kinetic energy depends on the square of the speed, so the speed at point C will be greater than the speed at point B by a factor of 2, not a factor of 2.

10.

(a)

The speed at the bottom of the hill does not depend on the angle of the hill if there is no friction. If there is no friction, then gravity is the only force doing work on the sled, and the system is conservative. All of the gravitational potential energy of the sled at the top of the hill will be converted into kinetic energy. The speed at the bottom of the hill depends on only the initial height h, not on the angle of the hill.

(b)

KEf

= 12 mu 2 = mgh, and u = (2 gh) 1/2.

The speed at the bottom of the hill does depend on the angle of the hill if there is friction. If friction is present, then the net force doing work on the sled is not conservative. Only part of the gravitational potential energy of the sled at the top of the hill will be converted into kinetic energy; the rest will be dissipated by the frictional force. The frictional force is proportional to the normal force on the sled, which will depend on the angle θ of the hill.

KEf

= 12 mu 2

= mgh - fx = mgh - µ mgh cos q / sin q = mgh (1 - µ / tan q ), so u = [2gh(1 - µ /tanq )]1/2 , which does depend on the angle of the hill and will be smaller for smaller angles.

11.

At the top of the pendulum’s swing, all of its energy is gravitational potential energy; at the bottom of the swing, all of the energy is kinetic. (a) If we can ignore friction, then energy is transformed back and forth between potential energy and kinetic energy as the pendulum swings. (b)

If friction is present, then during each swing energy is lost to friction at the pivot point and also to air resistance. During each swing, the kinetic energy and the potential energy decrease, and the pendulum’s amplitude decreases. When a grandfather clock is “wound up,” the amount of energy that will eventually be lost to friction and air resistance is stored as potential energy (either elastic or gravitational, depending on the clock mechanism), and part of the workings of the clock is to put that stored energy back into the pendulum at the same rate that friction is dissipating the energy.

12.

For each of the balloons, the initial energy (kinetic plus potential) equals the final energy (all kinetic). Since the initial energy depends on only the speed and not on the direction of the initial velocity, and all balloons have the same initial speed and height, the final speeds are all the same.

13.

The initial potential energy of the water is converted first into the kinetic energy of the water as it falls. When the falling water hits the pool, it does work on the water already in the pool, creating splashes and waves. Additionally, some energy is converted into heat and sound.

17.

The faster arrow has the same mass and twice the speed of the slower arrow, so the faster arrow will have four times the kinetic energy

(

2 KE = 1 mu 2

). Therefore, four times as much work must be done on the faster

arrow to bring it to rest. If the force on the arrows is constant, the faster arrow will travel four times the distance of the slower arrow into the hay.

Responses to MisConceptual Questions 1.

(b)

A common misconception is that all forces do work. However, work requires that the object on which the force is acting has a component of motion in the direction of the force.

2.

(c)

Work is done when the force acting on the object has a component in the direction of motion. Gravity, which provides the centripetal force, is not zero but is always perpendicular to the motion. A common error is the notion that an object moving in a circle has no work done on it. This is true only if the object is moving at constant speed.

3.

(c)

The kinetic energy is proportional to the square of the speed. Therefore, doubling the speed quadruples the kinetic energy.

4.

(d)

A common misconception is that the stopping distance is proportional to the speed. However, for a constant stopping force, the stopping distance is proportional to the initial kinetic energy, which is proportional to the square of the speed. Doubling the initial speed will quadruple the initial kinetic energy and therefore quadruple the stopping distance.

5.

(c)

As the ball falls, gravitational potential energy is converted to kinetic energy. As the ball hits the trampoline, kinetic energy is converted to elastic potential energy. This energy is then transferred back to kinetic energy of the ball and finally gravitational potential energy. No energy is added to the ball during the motion, so it can’t rise higher than it started. Some energy may be lost to heat during the motion, so the ball may not rise as high as it initially started.

6.

(e)

The term “energy” is commonly misunderstood. In this problem energy refers to the sum of the kinetic and potential energies. Initially the kinetic energy is a maximum. During the flight kinetic energy is converted to potential energy, with the potential energy a maximum at the highest point. As the ball falls back down the potential energy is converted back into kinetic energy. Since no nonconservative forces act on the ball (there is no air resistance), the total energy remains constant throughout the flight.

7.

(b)

Since the changes in speed are equal, many students think that the change in energy will also be equal. However, the energy is proportional to the square of the speed. It takes four times as much energy to accelerate the car from rest to 60 km/h as it takes to accelerate the car from rest to 30 km/h. Therefore, it takes three times the energy to accelerate the car from 30 km/h to 60 km/h as it takes to accelerate it from rest to 30 km/h.

12.

(c)

The kinetic energy depends on the speed and not the position of the block. Since the block moves with constant speed, the kinetic energy remains constant. As the block moves up the incline its elevation increases, so its potential energy also increases.

13.

(a)

The speed of the crate is constant, so the net (total) work done on the crate is zero. The normal force is perpendicular to the direction of motion, so it does no work. Your applied force and a component of gravity are in the direction of motion, so both do positive work. The force of friction opposes the direction of motion and does negative work. For the total work to be zero, the work done by friction must equal the sum of the work done by gravity and by you.

14.

(a)

The kinetic energy is proportional to the square of the ball’s speed, and the potential energy is proportional to the height of the ball. As the ball rises, the speed and kinetic energy decrease while the potential energy increases. As the ball falls, the speed and kinetic energy increase while the potential energy decreases.

Solutions to Problems 1.

The minimum force required to lift the firefighter is equal to his weight. The force and the displacement are both upward, so the angle between them is 0°. Use Eq. 6–1. W climb = Fclimbd cos q = mgd cos q = (75.0 kg)(9.80 m/s 2 )(28.0 m)cos 0 ° = 2.06 ´10 4 J

3.

Draw a free-body diagram for the crate as it is being pushed across the floor. Since it is not accelerating vertically, FN = mg . Since it is not accelerating horizontally, FP = Ffr = µ k F N = µ kmg . The work done to move it across the floor is the work done by the pushing force. The angle between the pushing force and the direction of motion is 0°.

! Ffr

! FP

! Dx

! FN

W push = F pushd cos 0° = µ kmgd (1) = (0.50)(46.0 kg)(9.80 m/s 2)(10.3 m) = 2300 J

4.

(a)

(b)

See the free-body diagram for the crate as it is being pulled. Since the crate is not accelerating horizontally, FP = Ffr = 230 N. The work done to move it across the floor is the work done by the pulling force. The angle between the pulling force and the direction of motion is 0°. Use Eq. 6–1. See the free-body diagram for the crate as it is being lifted. Since the crate accelerating vertically, the pulling force is the same magnitude as the weight. angle between the pulling force and the direction of motion is 0°.

WP = FP d cos0° = mgd = (1200 N)(5.0 m) = 6.0 × 103 J

! Ffr

! Dx ! FP

! FN

! mg

! Dy

! is not FP The

! mg

! mg

13.

The work done will be the area under the Fx vs. x graph. (a)

From x = 0.0 to x = 10.0 m, the shape under the graph is trapezoidal. The area is W = (400 N) 12 (10 m + 4 m) = 2800 J

(b)

From x = 10.0 m to x = 15.0 m, the force is in the opposite direction from the direction of motion, so the work will be negative. Again, since the shape is trapezoidal, we find W = (- 200 N) 12 (5 m + 2 m) = -700 J

Thus the total work from x = 0.0 to x = 15.0 m is 2800 J - 700 J = 2100 J .

15.

Find the velocity from the kinetic energy, using Eq. 6–3. KE

16.

(a)

= 12 mu 2

Since

KE

® u =

= 12 mu2 , u = 2 KE/m and u µ

multiplied by a factor of (b)

Since

KE

2KE 2(6.21´10 -21 J) = = 484 m/s m 5.31´ 10-26

= 12 mu2 ,

KE

KE .

Thus if the kinetic energy is tripled, the speed will be

3.

µu2 . Thus if the speed is halved, the kinetic energy will be multiplied by a

factor of 1/ 4 . 17.

The work done on the electron is equal to the change in its kinetic energy, Eq. 6–4. W = DKE = 1 mu 22 - 1 mu12 = 0 - 1 (9.11 ´10- 31 kg)(1.10 ´10 6 m/s) 2 = -5.51 ´10 - 19 J 2

2

2

Note that the work is negative since the electron is slowing down. 18.

The work done on the car is equal to the change in its kinetic energy, Eq. 6–3. é æ 1 m/s W = DKE = 1 mu 22 - 1 mu12 = 0 - 1 (925 kg) ê(95 km/h) ç 2 2 2 è3.6 km/h ë

2

öù 5 ÷ú = -3.2 ´10 J øû

Note that the work is negative since the car is slowing down. 26.

Subtract the initial gravitational PE from the final gravitational PE. D PE G = mgy 2 - mgy1 = mg (y 2 - y1 ) = (54 kg)(9.80 m/s2 )(4.0 m) = 2100 J

27.

The potential energy of the spring is given by

PE el

= 12 kx 2 , where x is the distance of stretching or

compressing of the spring from its natural length. x=

28.

2PEel 2(45.0 J) = = 1.01 m 88.0 N/m k

The initial stretching is from the equilibrium position, x = 0. Use that to find the spring constant. PEinitial PEfinal

= 12 kx 2

® 6.0 J = 12 k (2.0 cm) 2

= 21 kx 2 = 21 (3)(6)2 = 54 J;

PEfinal

® k = 3.0 J/cm 2

- PEinitial = 48 J

31.

The forces on the skier are gravity and the normal force. The normal force is perpendicular to the direction of motion, so she does no work. Thus the skier’s mechanical energy is conserved. Subscript 1 represents the skier at the top of the hill, and subscript 2 represents the skier at the bottom of the hill. The ground is the zero location for gravitational potential energy (y = 0). We have u1 = 0, y1 = 285 m, and y2 = 0 (bottom of the hill). Solve for u 2 , the speed at the bottom, using Eq. 6–13. 1 2

mu12 + mgy1 = 12 mu22 + mgy2

® 0 + mgy1 =

1 2

mu22 + 0 ®

u 2 = 2 gy1 = 2(9.80 m/s 2)(105 m) = 45.4 m/s 32.

The only forces acting on Jane are gravity and the vine tension. The tension pulls in a centripetal direction, so can do no work—the tension force is perpendicular at all times to her motion. So Jane’s mechanical energy is conserved. Subscript 1 represents Jane at the point where she grabs the vine, and subscript 2 represents Jane at the highest point of her swing. The ground is the zero location for gravitational potential energy (y = 0). We have u1 = 5.0 m/s, y1 = 0, and u 2 = 0 (top of swing). Solve for y2 , the height of her swing. Use Eq. 6–13. 1 2

mu12 + mgy1 = 12 mu 22 + mgy2

y2 =

u12 2g

=

(5.0 m/s)2 2(9.80 m/s 2)

®

1 2

mu12 + 0 = 0 + mgy2

= 1.276 m » 1.3 m

®

v2 , y2 v1 , y1...