Question 1

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 2

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 3

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

Using conservation of energy, we can find the speed of the ball at the bottom of the swing. At the top of the swing, all of the energy is in the form of potential energy, so:

$mg\frac{L}{2}=mgh$

where $h$ is the height of the swing, $L$ is the length of the string, and $g$ is the acceleration due to gravity. Solving for $h$ gives:

$h=\frac{L}{2}-\frac{L}{2}\cos	heta=L\left(\frac{1}{2}-\frac{1}{2}\cos	hetaight)$

where $	heta$ is the angle the string makes with the vertical at the bottom of the swing. At the bottom of the swing, all of the energy is in the form of kinetic energy, so:

$\frac{1}{2}mv^2=mgh$

Substituting for $h$, we get:

$\frac{1}{2}mv^2=mgL\left(\frac{1}{2}-\frac{1}{2}\cos	hetaight)$

Solving for $v$, we get:

$v=\sqrt{gL(1-\cos	heta)}$

At the bottom of the swing, $\cos	heta=-1$, so:

$v=\sqrt{2gL}$

Plugging in numbers, we get:

$v=\sqrt{2	imes 9.81	imes 0.5}=3.13 \approx \boxed{	extbf{(C) }3.1 	ext{ m/s}}$

Question 4

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 5

A particle moves along the x axis under the influence of a stationary object. The net force on the particle, which is conservative, is given by F = (8N/m³)x³. If the potential energy is taken to be zero for x = 0 then the potential energy is given by:

Accepted Answer

The potential energy (U) associated with a conservative force is given by U = -∫F dx. Given F = (8N/m)x, integrating with respect to x gives U = -(8/2)x^2 = -4x^2. The correct unit for potential energy is Joules (J), so the answer is (-2 J/m^2)x^2, which corresponds to choice B.

Question 6

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 7

A projectile of mass 0.50 kg is fired with an initial speed of 10 m/s at an angle of 60  $\degree$ 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 8

The potential energy of a body of mass m is given by U = -mgx + 1/2kx². The corresponding force is:

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Question 9

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?

Accepted Answer

The smallest value of y such that the object will slide without losing contact with the track is when the normal force is zero at the top of the track. At this point, the only force acting on the object is gravity. Therefore, the centripetal force required to keep the object on the track is equal to the force of gravity. 
$m \frac{v^2}{R} = mg$
$v^2 = gR$
$v = \sqrt{gR}$
At the top of the track, the velocity of the object is zero (since it momentarily stops moving). Therefore, the total energy of the object is equal to its potential energy at the top of the track. 
$mg(2R-y) = \frac{1}{2}mv^2$
$mg(2R-y) = \frac{1}{2}m(gR)$
$2R-y = \frac{1}{2}R$
$y = \frac{3}{2}R$
However, this is the minimum height required for the object to make it around the loop. The question asks for the minimum height required for the object to make it around the loop without losing contact with the track. Therefore, we must subtract off the radius of the object to get the answer. 
$y_{min} = \frac{3}{2}R - R = \frac{1}{2}R$
Therefore, the answer is B.

Question 10

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 its unstretched potential energy, is:

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Question 11

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 12

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 13

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

The initial kinetic energy of the block is given by:
Kinitial = (1/2)mv^2 = (1/2)(0.5 kg)(2 m/s)^2 = 1 J

The final potential energy stored in the compressed spring is given by:
Ufinal = (1/2)kx^2
where x is the maximum compression of the spring.

Since energy is conserved, we have:
Kinitial = Ufinal

Substituting the values, we get:
1 J = (1/2)(800 N/m)x^2
x^2 = 0.0025 m^2 = (0.05 m)^2

Taking the square root, we get:
x = 0.05 m = 5 cm

Therefore, the maximum spring compression is 5 cm.

Question 14

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 tension in the cord when the object is at the lowest point of its swing?

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Question 15

The graphs below show the magnitude of the force on a particle as the particle moves along the positive x axis from the origin to x = x₁. The force is parallel to the x axis and is conservative. The maximum magnitude F₁ has the same value for all graphs. Rank the situations according to the change in the potential energy associated with the force, least (or most negative) to greatest (or most positive).

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Question 16

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 first step is to find the amplitude of the motion. From the given information, we know that when the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50 m/s. We can use the conservation of energy to find the amplitude of the oscillation.

At the equilibrium position, all the energy is stored in the spring as potential energy. At the position where the spring is 4.0 cm longer than its equilibrium length, all the energy is converted into kinetic energy. Therefore, we can write:

(1/2) kx^2 = (1/2) mv^2

where k is the spring constant, x is the amplitude, m is the mass of the block, and v is the speed of the block at the 4.0 cm displacement.

Plugging in the numbers, we get:

(1/2) (80 N/m) (0.04 m)^2 = (1/2) (0.50 kg) (0.50 m/s)^2

Solving for x, we get:

x = 0.068 m

The maximum speed occurs when the block is at the equilibrium position, so all the potential energy is converted into kinetic energy. Therefore, we can write:

(1/2) kx^2 = (1/2) mv^2

where k, x, and m are the same as before, and v is the maximum speed of the block.

Plugging in the numbers, we get:

(1/2) (80 N/m) (0.068 m)^2 = (1/2) (0.50 kg) v^2

Solving for v, we get:

v = 0.71 m/s

Therefore, the greatest speed of the block is 0.71 m/s, which corresponds to choice C.

Question 17

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 18

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 19

A small object slides along the frictionless loop-the-loop with a diameter of 3 m. What minimum speed must it have at the top of the loop in order to remain in contact with the loop?

Accepted Answer

At the top of the loop, the normal force is pointing downwards while gravity is still pointing downwards. For the object to remain in contact with the loop, the normal force must equal zero i.e. gravity must be equal to the centripetal force:
$$mg = \frac{mv^2}{r}$$
where $m$ is the mass of the object, $v$ is the velocity at the top, $r$ is the radius of the loop. In this case, $r=\frac{3}{2}$ m.
Solving for $v$:
$$v = \sqrt{gr} = \sqrt{(9.8	ext{ m/s}^2)\left(\frac{3}{2}	ext{ m}ight)}\approx 3.8	ext{ m/s}$$
Therefore, the minimum velocity required at the top of the loop is 3.8 m/s.

Question 20

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 block as it leaves the spring is:

Accepted Answer

The potential energy stored in the 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, so: 
$K = U_s = 0.049	ext{ J} \approx 4.9	ext{ J}$ 
Therefore, the answer is C.

Quiz 8: Potential Energy and Conservation of Energy

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

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:

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?

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:

A particle moves along the x axis under the influence of a stationary object. The net force on the particle, which is conservative, is given by F = (8N/m³) x³. If the potential energy is taken to be zero for x = 0 then the potential energy is given by:

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:

A projectile of mass 0.50 kg is fired with an initial speed of 10 m/s at an angle of 60 $\degree$ 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:

The potential energy of a body of mass m is given by U = -mgx + 1/2kx². The corresponding force is:

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?

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 its unstretched potential energy, is:

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:

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:

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:

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 tension in the cord when the object is at the lowest point of its swing?

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:

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:

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:

A small object slides along the frictionless loop-the-loop with a diameter of 3 m. What minimum speed must it have at the top of the loop in order to remain in contact with the loop?

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 block as it leaves the spring is:

Quiz 8: Potential Energy and Conservation of Energy

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

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:

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?

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:

A particle moves along the x axis under the influence of a stationary object. The net force on the particle, which is conservative, is given by F = (8N/m3) x3. If the potential energy is taken to be zero for x = 0 then the potential energy is given by:

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:

A projectile of mass 0.50 kg is fired with an initial speed of 10 m/s at an angle of 60 °\degree° 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:

The potential energy of a body of mass m is given by U = -mgx + 1/2kx2. The corresponding force is:

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?

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 its unstretched potential energy, is:

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:

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:

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:

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 tension in the cord when the object is at the lowest point of its swing?

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:

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:

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:

A small object slides along the frictionless loop-the-loop with a diameter of 3 m. What minimum speed must it have at the top of the loop in order to remain in contact with the loop?

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 block as it leaves the spring is:

A particle moves along the x axis under the influence of a stationary object. The net force on the particle, which is conservative, is given by F = (8N/m³) x³. If the potential energy is taken to be zero for x = 0 then the potential energy is given by:

A projectile of mass 0.50 kg is fired with an initial speed of 10 m/s at an angle of 60 $\degree$ 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:

The potential energy of a body of mass m is given by U = -mgx + 1/2kx². The corresponding force is: