Paul Peter Urone; Roger Hinrichs

Multiple Choice

9.1 Work, Power, and the Work–Energy Theorem

22 .

Which expression represents power?

$f d$
$m g h$
$\frac{mv^2}{2}$
$\frac{W}{t}$

23.

The work–energy theorem states that the change in the kinetic energy of an object is equal to what?

The work done on the object
The force applied to the object
The loss of the object’s potential energy
The object’s total mechanical energy minus its kinetic energy

24.

A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s . If the runner has a mass of 60 kg, how long did it take him to reach that speed?

0.33 s
0.83 s
1.2 s
3.0 s

25.

A car’s engine generates 100,000 W of power as it exerts a force of 10,000 N. How long does it take the car to travel 100 m?

0.001 s
0.01 s
10 s
1,000 s

9.2 Mechanical Energy and Conservation of Energy

26 .

Why is this expression for kinetic energy incorrect?

\text{KE} = (m)(v)^2

.

The constant $g$ is missing.
The term $v$ should not be squared.
The expression should be divided by $2$ .
The energy lost to friction has not been subtracted.

27 .

What is the kinetic energy of a

10\text{kg}

object moving at

2.0\,\text{m/s}

?

$10\,\text{J}$
$20\,\text{J}$
$40\,\text{J}$
$100\,\text{J}$

28 .

Which statement best describes the PE-KE transformations for a javelin, starting from the instant the javelin leaves the thrower's hand until it hits the ground.

Initial PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, KE is transformed into PE. At every point in the flight, mechanical energy is being transformed into heat energy.
Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, PE is transformed into KE. At every point in the flight, mechanical energy is being transformed into heat energy.
Initial PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy.
Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy.

29 .

At the beginning of a roller coaster ride, the roller coaster car has an initial energy mostly in the form of PE. Which statement explains why the fastest speeds of the car will be at the lowest points in the ride?

At the bottom of the slope kinetic energy is at its maximum value and potential energy is at its minimum value.
At the bottom of the slope potential energy is at its maximum value and kinetic energy is at its minimum value.
At the bottom of the slope both kinetic and potential energy reach their maximum values
At the bottom of the slope both kinetic and potential energy reach their minimum values.

9.3 Simple Machines

30.

A large radius divided by a small radius is the expression used to calculate the IMA of what?

A screw
A pulley
A wheel and axle
An inclined plane.

31 .

(credit: modification of work by Shona Leonard|/DocPlayer)

Study the given pulley system.

How much force is needed to lift an object with a mass of 50 kg using this system? (Take Earth's g as 10 m/s2.)

83 N.
100 N.
125 N.
133 N.

32.

Which statement correctly describes the simple machines, like the crank in the image, that make up an Archimedes screw and the forces it applies?

The crank is a wedge in which the IMA is the length of the tube divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force.
The crank is an inclined plane in which the IMA is the length of the tube divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force.
The crank is a wheel and axle. The effort force of the crank becomes the resistance force of the screw.
The crank is a wheel and axle. The resistance force of the crank becomes the effort force of the screw.

33.

Refer to the pulley system on right in the image. Assume this pulley system is an ideal machine.
How hard would you have to pull on the rope to lift a 120 N load?
How many meters of rope would you have to pull out of the system to lift the load 1 m?

A series of 3 pulley systems are shown in this image. At the top of the image there is a formula I M A equals N. The first pulley on the left is shown. A red arrow is following diagonally down and to the right from the top of the pulley labeled F sub e. A weight is shown on the left side of the pulley with a red arrow from the weight denoted F sub r. A dotted line goes across the line from the weight to the pulley that reads N equals 1. The pulley in the center has a red arrow on the right side of the pulley going directly up labeled F sub e. A dotted line goes between the pulley and the fixed portion labeled N equals 2. A weight is suspended straight down from the pulley. The pulley system on the right has four pulleys in vertical succession. The top pulley and bottom pulley are larger than the pulleys in the middle. A red arrow is following diagonally down and to the right from the top of the pulley labeled F sub e. A dotted line goes between the two smaller pulleys in the middle labeled N equals 4. A hook on the bottom of the second pulley from the top starts a line that winds under the second smaller pulley, over the first smaller pulley, under the bottom larger pulley and over the top larger pulley. A weight is suspended from the bottom of the larger pulley.

480 N
4 m
480 N
$\frac{1}{4}$ m
30 N
4 m
30 N
$\frac{1}{4}$ m