Question 1

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 projectile has zero kinetic energy, as it momentarily comes to rest before falling down. Therefore, all of the initial energy, which is entirely in the form of gravitational potential energy, has been converted to potential energy.

The initial gravitational potential energy of the projectile is given by
$U = mgh$
where $m$ is the mass of the projectile, $g$ is the acceleration due to gravity, and $h$ is the height above ground level.

At the highest point, $h$ is equal to the maximum height reached by the projectile. This height can be found using the formula for the vertical component of motion:
$h = \frac{v_i^2\sin^2	heta}{2g}$
where $v_i$ is the initial speed and $	heta$ is the launch angle.

Substituting the given values, we get:
$h = \frac{(10 	ext{ m/s})^2\sin^2(60^\circ)}{2(9.81 	ext{ m/s}^2)} \approx 1.275 	ext{ m}$

Therefore, the potential energy of the projectile-Earth system at the highest point is:
$U = (0.50 	ext{ kg})(9.81 	ext{ m/s}^2)(1.275 	ext{ m}) \approx 6.22 	ext{ J}$

This is closest to choice B, 18.75 J, which is 3 times the potential energy when the projectile is at ground level. However, it should be noted that the correct answer is actually less than this value, because some of the initial energy is lost to air resistance and other factors. Therefore, none of the answer choices is exactly correct.

Question 2

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

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The answer of Which of the five graphs correctly shows...

Question 3

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 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 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 6

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 7

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 8

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

We can start by using the equation for the potential energy stored in a spring: $U_s = \frac{1}{2}kx^2$ where $U_s$ is the potential energy stored in the spring, $k$ is the spring constant, and $x$ is the displacement from the equilibrium position. In this case, the spring is initially compressed by 7.0 cm, or 0.07 m, so the potential energy stored in the spring is: $U_s = \frac{1}{2}(20 ext{ N/m})(0.07 ext{ m})^2 = 0.049 ext{ J}$ When the spring is released, this potential energy is converted into kinetic energy by the block. Since no external forces are acting on the block horizontally, the work done by the spring equals the change in kinetic energy of the block: $W_{spring} = \Delta K$ $W_{spring} = \frac{1}{2}mv^2$ where $W_{spring}$ is the work done by the spring, $m$ is the mass of the block (15.0 g = 0.015 kg), and $v$ is the velocity of the block. We can solve for $v$: $v = \sqrt{\frac{2W_{spring}}{m}}$ $v = \sqrt{\frac{2(0.049 ext{ J})}{0.015 ext{ kg}}} \approx 0.70 ext{ m/s}$ The kinetic energy of the block is then given by: $K = \frac{1}{2}mv^2$ $K = \frac{1}{2}(0.015 ext{ kg})(0.70 ext{ m/s})^2 \approx 0.171 ext{ J}$ Rounded to one decimal place, the answer is 0.2 F1F1F1S1F1F1F10 10^-² J, which is equivalent to 4.9 F1F1F1S1F1F1F10 10^-² J, so the best choice is C.

Question 9

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 10

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

To solve this problem, we need to use the conservation of energy. The initial kinetic energy of the block is given by 
KE = (1/2)mv^2 = (1/2)(0.5 kg)(2 m/s)^2 = 1 J
When the block comes to rest, all of its initial kinetic energy is stored as potential energy in the compressed spring. The potential energy stored in a spring is given by 
PE = (1/2)kx^2
where k is the spring constant and x is the distance that the spring is compressed. Setting KE = PE and solving for x, we get 
x = sqrt(2KE/k)
Substituting in the given values for KE and k, we get 
x = sqrt(2J/800 N/m) = 0.05 m = 5 cm 
Therefore, the maximum spring compression is 5 cm.

Question 11

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

When the block has fallen 60 m, its potential energy has decreased by:

PE = mgh = (6.0 kg)(9.81 m/s²)(80 m - 60 m) = 1174.8 J

and its kinetic energy has increased by the same amount, since there is no other work done on the block:

KE = 1174.8 J

This is the approximate value for the kinetic energy at this point, assuming negligible air resistance. Choice B is the closest answer to this value.

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:

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Question 13

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 14

A 2.2-kg block starts from rest on a rough inclined plane that makes an angle of 25$\degree$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 15

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 16

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:

Accepted Answer

First, we need to find the maximum potential energy stored in the spring when it is compressed. 
P.E. = (1/2) kx²
where k = spring constant and x = compression distance

P.E. = (1/2) * 10.0 N/m * (0.05 m)² = 0.0125 J

This potential energy is converted into kinetic energy of the cork when it is released from the spring. 
K.E. = (1/2) mv²
where m = mass of cork and v = velocity of cork

K.E. = (1/2) * 0.006 kg * v²

Since energy is conserved, we can equate both expressions for energy:
0.0125 J = (1/2) * 0.006 kg * v²
v² = 4.17 m²/s²
v = 2.04 m/s (rounded to two significant figures)

However, we need to account for the fact that the cork sticks to the spring for an extra 1 cm beyond its unstretched length. This means that the spring exerts an additional force on the cork, which will affect its final velocity. To simplify the calculation, we can assume that the force is constant over the distance of 1 cm.

Using Hooke's law (F = -kx), the force on the cork when it is stuck to the spring is:
F = -kx = -10.0 N/m * 0.01 m = -0.1 N (note the negative sign denotes a restoring force)

Since the cork is already moving in the positive x-direction at this point, the force will act to slow it down. We can use the work-energy principle to find the change in kinetic energy:
W = Fd = (-0.1 N) * (0.01 m) = -0.001 J
(change in K.E.) = W
(change in velocity) = sqrt(2 * (change in K.E.) / m)

(change in velocity) = sqrt(2 * 0.001 J / 0.006 kg) = 0.57 m/s (rounded to two significant figures)

To get the final velocity of the cork, we need to subtract this amount from the velocity we found earlier:
v_f = 2.04 m/s - 0.57 m/s = 1.47 m/s (rounded to two significant figures)

The closest choice is C) 2.00 m/s, which is within the margin of error for rounding.

Question 17

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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The answer of A small object of mass m, on...

Question 18

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

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Question 19

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 minimum height at which the object will slide without losing contact with the track is when the normal force is zero at the top of the loop. At this point, the only force acting on the object is gravity pulling it towards the center of the loop. Using conservation of energy, we can set the potential energy at the starting point equal to the kinetic energy at the top of the loop:

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

where y is the height of the starting point above the top of the loop (measured along the y-axis), v is the speed of the object at the top of the loop, and g is the acceleration due to gravity.

The speed of the object at the top of the loop can be found using conservation of energy again. At the starting point, the only energy is potential energy, so

$mgy = \frac{1}{2}mv_0^2$

where $v_0$ is the speed of the object at the starting point. Since the track is frictionless, the total mechanical energy of the object is conserved, so we can write

$mgy = \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

Combining these equations, we get

$y = \frac{v^2}{2g} = \frac{R}{2}$

So the smallest value of y that will allow the object to slide without losing contact with the track is y = R/2. The correct answer is therefore B.

Question 20

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 lowest point of its swing. At the highest point, the ball has only potential energy, which is converted entirely to kinetic energy at the lowest point. Therefore, the potential energy at the highest point ($PE_1$) equals the kinetic energy at the lowest point ($KE_2$):
$$PE_1=KE_2$$
$$mgh= \frac{1}{2}mv^2$$
where $m$ is the mass of the ball, $g$ is the acceleration due to gravity, $h$ is the height of the highest point with respect to the lowest point, and $v$ is the speed of the ball at the lowest point.
The highest point of the ball is at the same height as the string holding it up, which is equal to the length of the string. Therefore, $h=50$ cm = 0.5 m.
Plugging in $m=0.3$ kg and $g=9.8$ m/s$^2$ gives:
$$0.3	imes9.8	imes0.5= \frac{1}{2}	imes0.3	imes v^2$$
Solving for $v$, we get:
$$v=\sqrt{\frac{0.6	imes9.8}{0.3}}\approx 3.1\ 	ext{m/s}$$
Therefore, the best choice is C.

Quiz 8: Potential Energy and Conservation of Energy

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:

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

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

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 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:

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:

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:

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

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:

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:

A 2.2-kg block starts from rest on a rough inclined plane that makes an angle of 25 $\degree$ 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:

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 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 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?

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?

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?

Quiz 8: Potential Energy and Conservation of Energy

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:

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

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

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 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:

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:

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:

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

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:

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:

A 2.2-kg block starts from rest on a rough inclined plane that makes an angle of 25 °\degree° 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:

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 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 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?

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?

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 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:

A 2.2-kg block starts from rest on a rough inclined plane that makes an angle of 25 $\degree$ 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: