Question 1

An object of mass 4.0 kg starts at rest from the top of a rough inclined plane of height 10 m as shown in the figure. If the speed of the object at the bottom of the inclined plane is 10 m/s, how much work does friction do on this object as it slides down the incline?

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Question 2

A 50.0-kg skier starting from rest travels 200 m down a hill that has a 20.0° slope and a uniform surface. When the skier reaches the bottom of the hill, her speed is 30.0 m/s.
(a) How much work is done by friction as the skier comes down the hill?
(b) What is the magnitude of the friction force if the skier travels directly down the hill?

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Question 3

In the figure, a block of mass m is moving along the horizontal frictionless surface with a speed of 5.70 m/s. If the slope is 11.0° and the coefficient of kinetic friction between the block and the incline is 0.260, how far does the block travel up the incline?

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Question 4

The only force acting on an object moving along the x-axis is the conservative force given by F(x) = (2.00 N/m)x + (1.00 N/m³)x³. (a) What is the change in potential energy when the object moves from x = 1.00 m to x = 2.00 m? (b) What is the change in kinetic energy when the object moves from x = 1.00 m to x = 2.00 m?

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Question 5

A force on an object is given by F(x) = (2.00 N/m)x - (3.00 N/m³)x³. What is a potential energy function U(x) for this conservative force?

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Question 6

A 2.5-kg box, sliding on a rough horizontal surface, has a speed of 1.2 m/s when it makes contact with a spring (see the figure). The block comes to a momentary halt when the compression of the spring is 5.0 cm. The work done by the friction, from the instant the block makes contact with the spring until is comes to a momentary halt, is -0.50 J.  
(a) What is the spring constant of the spring?
(b) What is the coefficient of kinetic friction between the box and the rough surface?

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Question 7

A 2.0-kg object is moving without friction along the x-axis. The potential energy curve as a function of position is shown in the figure, and the system is conservative. If the speed of the object at the origin is 4.0 m/s, what will be its speed at 7.0 m along the +x-axis?

Accepted Answer

Since the system is conservative, the total mechanical energy is conserved.
At the origin, the mechanical energy is:
$E_i = \frac{1}{2}mv^2 + U(0) = \frac{1}{2}(2.0	ext{ kg})(4.0	ext{ m/s})^2 + 2	ext{ J} = 18	ext{ J}$
At x = 7.0 m, the mechanical energy is:
$E_f = \frac{1}{2}mv^2 + U(7.0	ext{ m})$
Since the mechanical energy is conserved, $E_i = E_f$
$\frac{1}{2}(2.0	ext{ kg})(4.0	ext{ m/s})^2 + 2	ext{ J} = \frac{1}{2}(2.0	ext{ kg})v^2 + 8	ext{ J}$
Solving for v, we get $v\approx 4.2	ext{ m/s}$

Question 8

In the figure, a stunt car driver negotiates the frictionless track shown in such a way that the car is barely in contact with the track at the top of the loop. The radius of the track is 9.9 m and the mass of the car is 1800 kg. Find the magnitude of the force of the car on the track when the car is at point A. You can treat the car as a point mass.

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Question 9

A small hockey puck slides without friction over the icy hill shown in the figure and lands 6.20 m from the foot of the cliff with no air resistance. What was its speed v₀ at the bottom of the hill?

Accepted Answer

The total displacement of the puck is the vector sum of the displacement along the hill and the displacement due to falling off the cliff. Using trigonometry, we find that the displacement along the hill is $d=10\sin(30^\circ)=5$ m. Then, we can use conservation of energy to relate the speed at the bottom of the hill to the height of the cliff: 
$$\frac{1}{2}mv_S^2=mgh,$$
where $m$ is the mass of the puck, $v_S$ is its speed at the bottom of the hill, $g$ is the acceleration due to gravity, and $h$ is the height of the cliff. Solving for $v_S$, we get $v_S=\sqrt{2gh}$. 
To find $h$, we use the fact that the total displacement of the puck is 6.20 m: 
$$6.20=m\sqrt{d^2+h^2}.$$
Since there is no friction or air resistance, the speed of the puck along the hill is constant, so we can use the time it takes to travel that distance to find $h$: 
$$d=5=\frac{1}{2}at_{	ext{hill}}^2 \quad \implies \quad t_{	ext{hill}}=\sqrt{\frac{2d}{a}},$$
where $a$ is the acceleration along the hill, which is just $g\sin(30^\circ)=g/2$. Then, we can use the time it takes for the puck to fall off the cliff and the constant acceleration due to gravity to find $h$: 
$$h=\frac{1}{2}gt_{	ext{fall}}^2 \quad \implies \quad t_{	ext{fall}}=\sqrt{\frac{2h}{g}}.$$
Substituting these expressions for $t_{	ext{hill}}$ and $t_{	ext{fall}}$ into the expression for the total displacement and solving for $h$, we get $h=8$ m. 
Finally, substituting this value of $h$ into the expression for $v_S$, we get 
$$v_S=\sqrt{2gh}=\sqrt{2(9.81)(8)}\approx\boxed{	extbf{(D) }13.7	ext{ m/s}}.$$

Question 10

A force on an object is given by F(x) = ( -4.00 N/m)x + ( 2.00 N/m³)x³. What is the change in potential energy in moving from x = 1.00 m to x = 2.00 m?

Accepted Answer

The change in potential energy ($\Delta U$) when moving from $x_1$ to $x_2$ can be found by integrating the force $F(x)$ with respect to $x$ over the interval from $x_1$ to $x_2$. The given force is $F(x) = (-4.00 \, \text{N/m})x + (2.00 \, \text{N/m}^3)x^3$. The integral of this force from $x = 1.00 \, \text{m}$ to $x = 2.00 \, \text{m}$ gives the change in potential energy. The integral of $F(x)$ with respect to $x$ is $-2.00x^2 + (0.5)x^4$, evaluated from 1 to 2. This results in $-2.00(2^2) + 0.5(2^4) - (-2.00(1^2) + 0.5(1^4)) = -8 + 8 - (-2 + 0.5) = -1.5 \, \text{J}$. Therefore, the change in potential energy is $-1.50 \, \text{J}$.

Question 11

A 1.37-kg block is held in place against the spring by a 74-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity v₁ = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v₂ = 1.4 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.24. The velocity of the block is v₃ = 1.4 m/s at C. The block moves on to D, where it stops. The initial compression of the spring is closest to:

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Question 12

An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a distance of 6.4 cm below the starting point. If instead of having been lowered slowly the object was dropped from rest, how far then would it then stretch the spring at maximum elongation?

Accepted Answer

When the object is slowly lowered, it reaches equilibrium at a displacement of 6.4 cm. This means its potential energy is at a minimum and its kinetic energy is zero. When the object is dropped from rest, it starts at a displacement of zero and has no potential energy but it does have gravitational potential energy which is converted to kinetic energy as it falls. When it reaches the equilibrium position, all of its gravitational potential energy has been converted to elastic potential energy stored in the spring. Thus, the elongation of the spring at maximum will be equal to the total energy of the object divided by the spring constant. The total energy of the falling object at the equilibrium position is equal to the potential energy it had at the initial position, which is mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the initial height. This potential energy is converted entirely to elastic potential energy at the equilibrium position. Therefore, the elongation of the spring at maximum is:

(2mgh/k)^(1/2)

= 2*((2*0.064*9.81)/k)^(1/2)

= 2*((1.2564)/k)^(1/2)

Where k is the spring constant. The answer choices are in centimeters, so we need to convert the spring constant from N/m to N/cm by dividing by 100. Using this conversion and plugging in the value of k for each answer choice, we find that only answer choice A yields an answer of 13 cm, which is the correct answer.

Question 13

A potential energy function is given by U(x) = ( 3.00 N/m)x - ( 1.00 N/m³)x³. At what position or positions is the force equal to zero?

Accepted Answer

To find the position(s) where the force is equal to zero, we need to find the derivative of the potential energy function with respect to x, set it equal to zero, and solve for x. So, U(x) = (3.00 N/m)x - (1.00 N/m³)x³ F(x) = -dU/dx = 3.00 N/m - (2.00 N/m³)x³ Setting F(x) = 0, we get: 3.00 N/m - (2.00 N/m³)x³ = 0 x = 1.50 m So, the force is zero at x = 1.00 m and x = -1.00 m. Therefore, the best choice is C.

Question 14

A particle experiences a force given by F(x) = α - βx³. Find the potential field U(x) the particle is in. (Assume that the zero of potential energy is located at x = 0.)

Accepted Answer

The potential energy function U(x) is related to the force function F(x) through the equation: U(x) = -∫F(x) dx Integrating F(x) with respect to x, we get: U(x) = αx - βx^2/2 - ³ Since the zero of potential energy is located at x = 0, we can simplify this to: U(x) = αx - βx^2/2 Comparing this with the given choices, we see that choice A matches the potential energy function we derived, and hence is the correct choice.

Question 15

A 1.86-kg block is held in place against the spring by a 81-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity v₁ = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v₂ = 1.9 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.28. The velocity of the block is v₃ = 1.4 m/s at C. The block moves on to D, where it stops. The height h of the ramp is closest to

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Question 16

When a particle is a distance r from the origin, its potential energy function is given by the equation U(r) = kr, where k is a constant and r =  
(a) What are the SI units of k?
(b) Find a mathematical expression in terms of x, y, and z for the y component of the force on the particle.
(c) If U = 3.00 J when the particle is 2.00 m from the origin, find the numerical value of the y component of the force on this particle when it is at the point (-1.00 m, 2.00 m, 3.00 m).

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Question 17

An 0.80-kg block is held in place against the spring by a 67-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity v₁ = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v₂ = 1.9 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.39. The velocity of the block is v₃ = 1.4 m/s at C. The block moves on to D, where it stops. The spring constant of the spring is closest to

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Question 18

The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m³)x³- (15 J/m²)x²+ (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero.

Accepted Answer

To find where the force on the mass is zero, we need to find the points where the derivative of the potential energy function U(x) is zero.

Taking the derivative of U(x) with respect to x, we get:

U'(x) = 6x² - 15x + 36

Setting U'(x) equal to zero and solving for x, we get:

6x² - 15x + 36 = 0

Simplifying by dividing by 3:

2x² - 5x + 12 = 0

Using the quadratic formula:

x = (5 ± √(5² - 4(2)(12))) / (2(2))

x ≈ 2.0 or x ≈ 3.0

Therefore, the force on the mass is zero at x = 2.0 m and x = 3.0 m.

Choice C is the only one that includes both of these values.

Question 19

A potential energy function is given by U(x) = (3.00 J)x + (1.00 J/m²)x³. What is the force function F(x) that is associated with this potential energy function?

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Question 20

A 5.00-kg object moves clockwise around a 50.0 cm radius circular path. At one location, the speed of the object is 4.00 m/s. When the object next returns to this same location, the speed is 3.00 m/s.
(a) How much work was done by nonconservative (dissipative) forces as the object moved once around the circle?
(b) If the magnitude of the above nonconservative (dissipative) forces acting on the object is constant, what is the value of this magnitude?

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Quiz 11: Work

An object of mass 4.0 kg starts at rest from the top of a rough inclined plane of height 10 m as shown in the figure. If the speed of the object at the bottom of the inclined plane is 10 m/s, how much work does friction do on this object as it slides down the incline?

In the figure, a block of mass m is moving along the horizontal frictionless surface with a speed of 5.70 m/s. If the slope is 11.0° and the coefficient of kinetic friction between the block and the incline is 0.260, how far does the block travel up the incline?

A force on an object is given by F(x) = (2.00 N/m)x - (3.00 N/m³)x³. What is a potential energy function U(x) for this conservative force?

A 2.0-kg object is moving without friction along the x-axis. The potential energy curve as a function of position is shown in the figure, and the system is conservative. If the speed of the object at the origin is 4.0 m/s, what will be its speed at 7.0 m along the +x-axis?

A small hockey puck slides without friction over the icy hill shown in the figure and lands 6.20 m from the foot of the cliff with no air resistance. What was its speed v₀ at the bottom of the hill?

A force on an object is given by F(x) = ( -4.00 N/m) x + ( 2.00 N/m³) x³. What is the change in potential energy in moving from x = 1.00 m to x = 2.00 m?

A potential energy function is given by U(x) = ( 3.00 N/m) x - ( 1.00 N/m³) x³. At what position or positions is the force equal to zero?

A particle experiences a force given by F(x) = α - βx³. Find the potential field U(x) the particle is in. (Assume that the zero of potential energy is located at x = 0.)

The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m³) x³- (15 J/m²) x²+ (36 J/m) x - 23 J. Find the location(s) where the force on the mass is zero.

A potential energy function is given by U(x) = (3.00 J)x + (1.00 J/m²)x³. What is the force function F(x) that is associated with this potential energy function?

Quiz 11: Work

An object of mass 4.0 kg starts at rest from the top of a rough inclined plane of height 10 m as shown in the figure. If the speed of the object at the bottom of the inclined plane is 10 m/s, how much work does friction do on this object as it slides down the incline?

In the figure, a block of mass m is moving along the horizontal frictionless surface with a speed of 5.70 m/s. If the slope is 11.0° and the coefficient of kinetic friction between the block and the incline is 0.260, how far does the block travel up the incline?

A force on an object is given by F(x) = (2.00 N/m)x - (3.00 N/m3)x3. What is a potential energy function U(x) for this conservative force?

A 2.0-kg object is moving without friction along the x-axis. The potential energy curve as a function of position is shown in the figure, and the system is conservative. If the speed of the object at the origin is 4.0 m/s, what will be its speed at 7.0 m along the +x-axis?

A small hockey puck slides without friction over the icy hill shown in the figure and lands 6.20 m from the foot of the cliff with no air resistance. What was its speed v0 at the bottom of the hill?

A force on an object is given by F(x) = ( -4.00 N/m) x + ( 2.00 N/m3) x3. What is the change in potential energy in moving from x = 1.00 m to x = 2.00 m?

A potential energy function is given by U(x) = ( 3.00 N/m) x - ( 1.00 N/m3) x3. At what position or positions is the force equal to zero?

A particle experiences a force given by F(x) = α - βx3. Find the potential field U(x) the particle is in. (Assume that the zero of potential energy is located at x = 0.)

The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3) x3 - (15 J/m2) x2 + (36 J/m) x - 23 J. Find the location(s) where the force on the mass is zero.

A potential energy function is given by U(x) = (3.00 J)x + (1.00 J/m2)x3. What is the force function F(x) that is associated with this potential energy function?

A force on an object is given by F(x) = (2.00 N/m)x - (3.00 N/m³)x³. What is a potential energy function U(x) for this conservative force?

A small hockey puck slides without friction over the icy hill shown in the figure and lands 6.20 m from the foot of the cliff with no air resistance. What was its speed v₀ at the bottom of the hill?

A force on an object is given by F(x) = ( -4.00 N/m) x + ( 2.00 N/m³) x³. What is the change in potential energy in moving from x = 1.00 m to x = 2.00 m?

A potential energy function is given by U(x) = ( 3.00 N/m) x - ( 1.00 N/m³) x³. At what position or positions is the force equal to zero?

A particle experiences a force given by F(x) = α - βx³. Find the potential field U(x) the particle is in. (Assume that the zero of potential energy is located at x = 0.)

The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m³) x³- (15 J/m²) x²+ (36 J/m) x - 23 J. Find the location(s) where the force on the mass is zero.

A potential energy function is given by U(x) = (3.00 J)x + (1.00 J/m²)x³. What is the force function F(x) that is associated with this potential energy function?