Question 1

The unit of power, watt, is dimensionally the same as:

Accepted Answer

Watt is defined as the rate of energy transfer, which is the amount of energy transferred per unit time. This can be expressed as joule/second, making (B) the correct choice. 
(A) is incorrect because it is the dimension of energy, not power. 
(C) and (D) are incorrect because they are the dimensions of work and work per unit length, respectively, not power.

Question 2

A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5.0 m/s. The hill is 10 m high and 100 m long. If the force of friction as she rides down the hill is 24 N, what is her speed at the bottom?

Accepted Answer

The answer of A girl and her bicycle have a...

Question 3

A 20-N crate starting at rest slides down a rough 5.0-m long ramp, inclined at 25 $\degree$ with the horizontal. 30 J of energy is lost to friction. What will be the speed of the crate at the bottom of the incline?

Accepted Answer

The gravitational potential energy lost is mgh = 20 N * 5 m * sin(25°). Subtract the 30 J lost to friction to find the kinetic energy at the bottom, then solve for speed using KE = 0.5 * m * v^2.

Question 4

How fast would a 1000-kg car be traveling if it had a kinetic energy equal to 1000 Calories?

Accepted Answer

The answer of How fast would a 1000-kg car be...

Question 5

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 800 m. The parachute deploys, and she lands on the ground with a speed of 5.0 m/s. How much energy was lost to air friction during this jump?

Accepted Answer

First, we can calculate the potential energy of the parachutist before jumping:

PE = mgh = (50.0 kg)(9.81 m/s^2)(800 m) = 392,400 J

Next, we can calculate the kinetic energy of the parachutist just before landing:

KE = (1/2)mv^2 = (1/2)(50.0 kg)(5.0 m/s)^2 = 625 J

The energy lost to air friction is then the difference between the initial potential energy and final kinetic energy:

Elost = PE - KE = 392,400 J - 625 J = 391,775 J

Rounding this to two significant figures gives us the answer of 391,000 J, which corresponds to choice C.

Question 6

The quantity of work equal to one joule is also equivalent to which of the following?

Accepted Answer

The answer of The quantity of work equal to one...

Question 7

Yuri, a Russian weightlifter, is able to lift 300 kg a distance 2.00 m in 2.00 s. What is his power output?

Accepted Answer

Power = work / time
Work = force x distance
Yuri lifted 300 kg (2940 N) a distance of 2 m, so his work was:
Work = 2940 N x 2 m = 5880 J
His time was 2 seconds. Therefore:
Power = 5880 J / 2 s = 2940 W = 2.94 kW

Question 8

A baseball catcher puts on an exhibition by catching a 0.150-kg ball dropped from a helicopter at a height of 100 m above the catcher. If the catcher "gives" with the ball for a distance of 0.500 m while catching it, what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

Accepted Answer

The answer of A baseball catcher puts on an exhibition...

Question 9

A 66-kg woman runs up a flight of stairs having a rise of 4.0 m in a time of 4.2 s. What average power did she supply?

Accepted Answer

First, we need to calculate the gravitational potential energy that the woman gained, which is given by mgh, where m is the mass of the woman, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the stairs. So, the potential energy gained is:
mgh = (66 kg)(9.8 m/s^2)(4.0 m) = 2587.2 J

Next, we can calculate the average power that the woman supplied using the formula:
P = E/t
where P is power, E is energy, and t is time. So, the average power supplied is:
P = 2587.2 J / 4.2 s = 616.57 W

Therefore, the best choice is C, which is closest to our calculated value of 616.57 W.

Question 10

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position. Under double the compression, the spring is now used to launch the mass. How high does the mass now rise above its starting position?

Accepted Answer

The answer of A Hooke's law spring is compressed a...

Question 11

I drop a 60-g golf ball from 2.0 m high. It rebounds to 1.3 m. How much energy is lost?

Accepted Answer

The answer of I drop a 60-g golf ball from...

Question 12

An automobile delivers 30.0 hp to its wheels when moving at a constant speed of 26.0 m/s. What is the resistance force on the automobile at this speed? (1 hp = 746 watts)

Accepted Answer

We can start by using the formula for power:
P = Fv

where P is power, F is force, and v is velocity.

Rearranging this formula to solve for force, we get:
F = P/v

Substituting the given values, we get:
F = (30.0 hp x 746 W/hp) / 26.0 m/s
F = 861 N

Therefore, the resistance force on the automobile at this speed is 861 N, which is answer choice D.

Question 13

Adisa pulls a 50-N crate up a 5.0-m long inclined plane at a constant velocity. If the plane is inclined at an angle of 37 $\degree$ to the horizontal and there is a constant force of friction of 10 N between the crate and the surface, what is the net change in potential energy of the crate?

Accepted Answer

The answer of Adisa pulls a 50-N crate up a...

Question 14

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position. Under the same compression d, the spring is now used to launch a mass of 2m. How high does this second mass rise?

Accepted Answer

The answer of A Hooke's law spring is compressed a...

Question 15

A pile driver drives a post into the ground. The mass of the pile driver is 2500 kg, and it is dropped through a height of 8.0 m on each stroke. If the resisting force of the ground is 2.0 * 10⁶ N, how far is the post driven in on each stroke?

Accepted Answer

The answer of A pile driver drives a post into...

Question 16

The rate at which work is done is equivalent to which of the following?

Accepted Answer

The rate at which work is done is defined as power, which is measured in watts (W). The other choices (increase in potential energy, thermal energy, and potential energy) are all forms of energy, but do not directly relate to the rate at which work is done.

Question 17

A block is projected with speed v across a horizontal surface and slides to a stop due to friction. The same block is then projected with the same speed v up an incline where it slides to a stop due to friction. In which case did the total mechanical energy of the block decrease the least?

Accepted Answer

The work done by friction is less on the incline because the normal force is less than on the horizontal surface.

Question 18

A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 12.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction?

Accepted Answer

First, we can find the total potential energy at the top of the hill:
PE = mgh = (40.0 kg)(9.8 m/s^2)(12.0 m) = 4704 J
Next, we can find the total kinetic energy at the bottom of the hill:
KE = (1/2)mv^2 = (1/2)(40.0 kg)(2*5.0 m/s)^2 = 1000 J
The difference between the initial potential energy and the final kinetic energy is the work done by friction, which converts energy into heat:
PE(initial) - KE(final) = 4704 J - 1000 J = 3704 J
Therefore, the amount of energy lost to friction is 3704 J. However, the question asks for the amount of kinetic energy and potential energy lost to friction. Since kinetic energy is lost, the answer is not A or C. Therefore, the correct answer is B) 3200 J, since it is closest to the calculated value.

Question 19

A Hooke's law spring is compressed a distance d and is used to launch a particle of mass m vertically to a height h above its starting position. Under double the compression, the spring is now used to launch a particle of mass 2 m. How high does the second mass rise above its starting position?

Accepted Answer

According to Hooke's law, the force exerted by a spring is proportional to its displacement. Therefore, if the spring is compressed twice as much, it will exert twice the force.

For the first launch, the force exerted by the spring is F = kx, where k is the spring constant and x is the distance the spring is compressed. This force is equal to the weight of the particle, mg, where g is the acceleration due to gravity. Therefore:

kx = mg

Solving for x:

x = mg/k

The work done by the spring is equal to the potential energy gained by the particle:

W = 1/2 kx^2 = 1/2 mg^2/k

This potential energy is converted into kinetic energy at the moment the particle leaves the spring, and then into potential energy as the particle rises to height h:

1/2 mv^2 = 1/2 mg^2/k

v = sqrt(2g^2/k)

h = v^2/2g = mg/k

For the second launch, the spring is compressed twice as much, so the force exerted by the spring will be twice as much. The particle has twice the mass, so the acceleration due to gravity will be the same as before. Therefore, the particle will rise to a height equal to:

2mg/(2k) = mg/k

This is the same height as before, so the correct answer is B) 2h.

Question 20

Preston pushes a wheelbarrow weighing 500 N to the top of a 50.0-m ramp, inclined at 20.0 $\degree$ with the horizontal and leaves it. Tamara accidentally bumps the wheelbarrow. It slides back down the ramp, during which a 60.0-N frictional force acts on it over the 50.0 m. What is the wheelbarrow's kinetic energy at the bottom at of the ramp? (g = 9.8 m/s²)

Accepted Answer

The gravitational potential energy at the top is converted into kinetic energy at the bottom minus the work done against friction. Potential energy = mgh = 500 N * 50 m * sin(20°) = 1710 J. Work done against friction = 60 N * 50 m = 3000 J. Kinetic energy = 1710 J - 3000 J = 5550 J.

Quiz 6: Work and Energy