Question 1

A simple pendulum consists of a 2.0 kg mass attached to a string. It is released from rest at X as shown. Its speed at the lowest point Y is:

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Question 2

A 2.2-kg block starts from rest on a rough inclined plane that makes an angle of 25 °  with the horizontal. The coefficient of kinetic friction is 0.25. As the block goes 2.0 m down the plane, the mechanical energy of the Earth-block system changes by:

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Question 3

An ideal spring is used to fire a 15.0-g block horizontally. The spring has a spring constant of 20 N/m and is initially compressed by 7.0 cm. The kinetic energy of the pellet as it leaves the spring is:

Accepted Answer

The potential energy stored in the compressed spring is:
$U_s = \frac{1}{2}kx^2 = \frac{1}{2}(20\ 	ext{N/m})(0.07\ 	ext{m})^2 = 0.049\ 	ext{J}$

This potential energy is converted entirely into kinetic energy as the block leaves the spring. Therefore, the kinetic energy of the block is also 0.049 J, which is equivalent to 4.9 J when rounded to one significant figure. Thus, the correct answer is C.

Question 4

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 0.12 J. The greatest speed of the block is:

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Question 5

The string in the figure is 50 cm long. When the ball is released from rest, it swings along the dotted arc. How fast is it going at the lowest point in its swing?

Accepted Answer

At point A (the release point), the ball has zero kinetic energy and zero potential energy. As it swings down to point B, it gains kinetic energy and loses potential energy. By the time it reaches point B, it has converted all of its potential energy to kinetic energy. Therefore, we can use conservation of energy to find the speed at point B.

Gravitational potential energy at A = mgh
Gravitational potential energy at B = 0

Therefore, the gain in kinetic energy = loss in potential energy
1/2mv^2 = mgh
v = sqrt(2gh)

Since the length of the string is 50 cm = 0.5 m, the height difference between A and B is h = 0.5 – 0.5cos(60) = 0.25 m

Plugging in the values, we get:
v = sqrt(2 x 9.8 x 0.25) = 3.14 m/s

Rounding to one significant figure, the answer is 3.1 m/s, which corresponds to choice C.

Question 6

A block of mass m is initially moving to the right on a horizontal frictionless surface at a speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy of the block is equal to the potential energy of the spring, the spring is compressed a distance of:

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Question 7

A 700-N man jumps out of a window into a fire net 10 m below. The net stretches 2 m before bringing the man to rest and tossing him back into the air. The maximum potential energy of the net, compared to it's unstretched potential energy, is:

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Question 8

A block is released from rest at point P and slides along the frictionless track shown. At point Q, its speed is:

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Question 9

A 6.0-kg block is released from rest 80 m above the ground. When it has fallen 60 m its kinetic energy is approximately:

Accepted Answer

The potential energy of the block at the height of 80 m can be calculated as:

PE = mgh = 6.0 kg x 9.81 m/s^2 x 80 m = 4708.8 J

When the block falls to a height of 60 m, its potential energy decreases and its kinetic energy increases. The work done by the gravitational force is converted into kinetic energy. Let's assume the velocity of the block when it has fallen 60 m is v.

The work done by the gravitational force is:

W = Fd = mgh - mgh' = ΔPE

where h' = 60 m and h = 80 m.

So, ΔPE = PE - PE' = 4708.8 J - mgh' = 4708.8 J - 6.0 kg x 9.81 m/s^2 x 60 m = 1765.28 J

This work done is equal to the change in kinetic energy:

ΔKE = KE - KE' = 1/2 mv^2 - 1/2 mv0^2 = 1/2 mv^2

where v0 = 0 (initially at rest).

Substituting ΔKE and solving for v gives:

ΔKE = 1/2 mv^2 = 1765.28 J

v = sqrt(2ΔKE/m) = sqrt(2 x 1765.28 J / 6.0 kg) = 28.87 m/s

The kinetic energy of the block when it has fallen 60 m can be calculated as:

KE = 1/2 mv^2 = 1/2 x 6.0 kg x (28.87 m/s)^2 = 2983.5 J

Thus, the kinetic energy of the block when it has fallen 60 m is approximately 3500 J (closest to B).

Question 10

A small object of mass m starts at rest at the position shown and slides along the frictionless loop-the-loop track of radius R. What is the smallest value of y such that the object will slide without losing contact with the track?

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Question 11

A rectangular block is moving along a frictionless path when it encounters the circular loop as shown. The block passes points 1,2,3,4,1 before returning to the horizontal track. At point 3:

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Question 12

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50 m/s. The greatest speed of the block is:

Accepted Answer

The greatest speed of the block occurs when the spring is at its equilibrium length. The potential energy stored in the spring at this point is equal to the kinetic energy of the block. Therefore, the greatest speed of the block is:$$v = \sqrt{\frac{2k}{m}} = \sqrt{\frac{2(80 N/m)}{0.50 kg}} = 0.78 m/s$$

Question 13

A small object of mass m, on the end of a light cord, is held horizontally at a distance r from a fixed support as shown. The object is then released. What is the in the cord when the object is at the lowest point of its swing?

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Question 14

The long pendulum shown is drawn aside until the ball has risen 0.5 m. It is then given an initial speed of 3.0 m/s.. The speed of the ball at its lowest position is:

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Question 15

Which of the five graphs correctly shows the potential energy of a spring as a function of its elongation x?

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Question 16

For a block of mass m to slide without friction up the rise of height h shown, it must have a minimum initial kinetic energy of:

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Question 17

A toy cork gun contains a spring whose spring constant is 10.0 N/m. The spring is compressed 5.00 cm and then used to propel a 6.00-g cork. The cork, however, sticks to the spring for 1.00 cm beyond its unstretched length before separation occurs. The muzzle velocity of this cork is:

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Question 18

A projectile of mass 0.50 kg is fired with an initial speed of 10 m/s at an angle of 60 ° above the horizontal. The potential energy of the projectile-Earth system when the projectile is at its highest point (relative to the potential energy when the projectile is at ground level) is:

Accepted Answer

At the highest point, the vertical component of the velocity is zero, and only the potential energy is considered. The initial kinetic energy can be used to find the maximum height using the equation $KE_i = PE_f$, where $KE_i = \frac{1}{2}mv^2_y$ and $PE_f = mgh$. The initial vertical velocity ($v_y$) can be found using $v_y = v\sin(\theta)$, where $v = 10 \, \text{m/s}$ and $\theta = 60^\circ$. Thus, $v_y = 10\sin(60^\circ) = 8.66 \, \text{m/s}$. The initial kinetic energy in the vertical direction is $KE_i = \frac{1}{2} \cdot 0.5 \cdot (8.66)^2 = 18.75 \, \text{J}$. This equals the potential energy at the highest point, so the correct answer is 18.75 J.

Question 19

An elevator is rising at constant speed. Consider the following statements: I. the upward cable force is constant
II. the kinetic energy of the elevator is constant
III. the gravitational potential energy of the Earth-elevator system is constant
IV. the acceleration of the elevator is zero
V. the mechanical energy of the Earth-elevator system is constant

Accepted Answer

The elevator is moving at a constant speed, so the upward cable force (I) is constant and equal to the gravitational force. The kinetic energy (II) is constant because speed is constant. The acceleration (IV) is zero because the speed is constant. The gravitational potential energy (III) is not constant as the elevator rises. The mechanical energy (V) is not constant because potential energy changes.

Question 20

A 0.5-kg block slides along a horizontal frictionless surface at 2 m/s. It is brought to rest by compressing a very long spring of spring constant 800 N/m. The maximum spring compression is:

Accepted Answer

Using conservation of energy, we can find the maximum compression of the spring. Initially, the block has kinetic energy of (1/2)mv^2 = (1/2)(0.5 kg)(2 m/s)^2 = 1 J. When the spring is fully compressed, all of the block's energy is stored as potential energy in the spring. That is, (1/2)kx^2 = 1 J, where k is the spring constant and x is the compression of the spring. Solving for x, we get x = sqrt(2/k) = sqrt(2/800) = 0.05 m = 5 cm. Therefore, the maximum spring compression is 5 cm.

Quiz 8: Potential Energy and Conservation of Energy