Question 1

The solution to a problem is the equation below.Which description best fits this solution? $$\frac { 1 } { 2 } \left( 500 \frac { \mathrm { N } } { \mathrm { m } } \right) ( 0.120 \mathrm {~m} ) ^ { 2 } - ( 2.00 \mathrm {~kg} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) ( 0.120 \mathrm {~m} )$$ $$= \frac { 1 } { 2 } ( 2.00 \mathrm {~kg} ) \left( 0.824 \frac { \mathrm { m } } { \mathrm { s } } \right) ^ { 2 } + ( 2.00 \mathrm {~kg} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) ( 0.029 0 \mathrm {~m} )$$
A)A vertical spring compressed 0.120 m shoots a 2.00 kg mass 2.90 cm above the equilibrium position of the spring.
B)A vertical spring stretched 0.120 m shoots a 2.00 kg mass 9.10 cm above the equilibrium position of the spring.
C)A vertical spring compressed 0.120 m shoots a 2.00 kg mass 12.0 cm above the equilibrium position of the spring.
D)A vertical spring compressed 0.120 m shoots a 2.00 kg mass 14.9 cm above the equilibrium position of the spring.
E)A 2.00 kg mass has fallen 0.820 m and compressed the upper end of a vertical spring 12.0 cm below the equilibrium position.

Accepted Answer

The equation represents the conservation of energy for a mass-spring system. The initial potential energy stored in the spring and gravitational potential energy is converted into kinetic energy and gravitational potential energy at a new height. The solution shows that the mass reaches a height of 0.0290 m (2.90 cm) above the equilibrium position, matching option A.

Question 2

A 3.0-kg mass is dropped from the edge of a 50-m tall building with an initial speed of zero.The mass strikes the ground with a downward velocity of 25 m/s.Find the change in mechanical energy of the mass caused by air resistance between the point where it is dropped and the point where it strikes the ground?
A)-0.46 kJ
B)-0.53 kJ
C)-0.61 kJ
D)-0.38 kJ
E)-0.81 kJ

Accepted Answer

The change in mechanical energy is equal to the work done by air resistance. We can calculate the potential energy of the mass before it falls using the equation: 
PE = mgh = (3.0 kg)(9.8 m/s^2)(50 m) = 1470 J

At the point of impact, the mass has lost all of its potential energy and gained kinetic energy: 
KE = (1/2)mv^2 = (1/2)(3.0 kg)(25 m/s)^2 = 937.5 J

Therefore, the change in mechanical energy is: 
ΔE = KE - PE = 937.5 J - 1470 J = -532.5 J

Since the mass has lost mechanical energy due to air resistance, the answer must be negative. The closest option is B: -0.53 kJ or -530 J, which represents a loss of 530 J of mechanical energy.

Question 3

Two masses,M_A and M_B,with M_B = 2M_A,are released at the same time and allowed to fall straight down.Neglect air resistance.When we compare their kinetic energies after they have fallen equal distances,we find that A)K_B = K_A. B)K_B = 2K_A. C)K_B = 4K_A. D)K_A = 2K_B. E)K_A = 4K_B.

Accepted Answer

The answer of Two masses,M_A and M_B,with M_B = 2M_A,are...

Question 4

A large spring is used to stop the cars after they come down the last hill of a roller coaster.The cars start at rest at the top of the hill and are caught by a mechanism at the instant their velocities at the bottom are zero.Compare the compression of the spring,x_A,for a fully loaded car with that,x_B,for a lightly loaded car when m_A = 2m_B. A)x_A = $\frac { 1 } { 2 }$ x_B. B)x_A = x_B. C)x_A = $\sqrt { 2 }$ x_B. D)x_A = 2 x_B. E)x_A = 4 x_B.

Accepted Answer

The answer of A large spring is used to stop...

Question 5

A 2.0-kg block slides down a fixed,rough curved track.The block has a speed of 5.0 m/s after its height above a horizontal surface has decreased by 1.8 m.Assume the block starts from rest.What is the change in mechanical energy of the block caused by the force of friction during this descent?
A)-14 J
B)-12 J
C)-10 J
D)-16 J
E)-25 J

Accepted Answer

The change in mechanical energy is equal to the work done by all forces acting on the object. In this case, the only force doing work is friction, which acts against the direction of motion, slowing the block down. The work done by friction is equal to the force of friction times the distance traveled, which can be calculated using the work-energy principle:

W_friction = ΔK = K_f - K_i

where W_friction is the work done by friction, ΔK is the change in kinetic energy, K_f is the final kinetic energy (when the block reaches the bottom of the track), and K_i is the initial kinetic energy (when the block starts from rest).

We can use the kinematic equations to find the final velocity of the block at the bottom of the track:

v_f^2 = v_i^2 + 2gh
v_f^2 = 0 + 2(9.8 m/s^2)(1.8 m)
v_f = 5.4 m/s

Then we can calculate the initial kinetic energy:

K_i = (1/2)mv_i^2 = 0

And the final kinetic energy:

K_f = (1/2)mv_f^2 = (1/2)(2.0 kg)(5.4 m/s)^2 = 29.16 J

Finally, we can calculate the work done by friction:

W_friction = K_f - K_i = 29.16 J - 0 J = 29.16 J

Therefore, the change in mechanical energy caused by the force of friction during the descent is -29.16 J (negative because friction does negative work). The closest answer choice to this value is choice C, which is 10 J off. None of the other choices are within a reasonable range.

Question 6

A 1.2-kg mass is projected up a rough circular track (radius = 0.80 m)as shown.The speed of the mass at point A is 8.4 m/s,and at point B,it is 5.6 m/s.What is the change in mechanical energy between A and B caused by the force of friction? 
A)-2.7 J
B)-8.8 J
C)-4.7 J
D)-6.7 J
E)-19 J

Accepted Answer

The answer of A 1.2-kg mass is projected up a...

Question 7

A 1.5-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 200 N/m)which has its other end fixed.If this system is displaced 20 cm horizontally from the equilibrium position and released from rest,the block first reaches the equilibrium position with a speed of 2.0 m/s.What is the coefficient of kinetic friction between the block and the horizontal surface on which it slides?
A)0.34
B)0.24
C)0.13
D)0.44
E)0.17

Accepted Answer

The answer of A 1.5-kg block sliding on a rough...

Question 8

A pendulum bob has potential energy U₀ when held taut in a horizontal position.The bob falls until it is 30$\degree$ away from the horizontal position,when it has potential energy U_A.It continues to fall until the string is vertical,when it has potential energy U_B.Compare its potential energies at O,A,and B. A)U₀ = U_A = U_B. B)U_A- U_B = 2U₀. C)U_A - U_B = U₀ - U_A. D)U₀ = U_B = 2U_A. E)U₀ - U_A = 2(U_A -U_B).

Accepted Answer

The answer of A pendulum bob has potential energy U₀...

Question 9

A 10-kg object is dropped from rest.After falling a distance of 50 m,it has a speed of 26 m/s.What is the change in mechanical energy caused by the dissipative (air)resistive force on the object during this descent?
A)-1.3 kJ
B)-1.5 kJ
C)-1.8 kJ
D)-2.0 kJ
E)-2.3 kJ

Accepted Answer

The answer of A 10-kg object is dropped from rest.After...

Question 10

A spring with spring constant k = 800 N/m is extended 12 cm from its equilibrium position.A spring with 6.0 cm extension from equilibrium will have the same potential energy as the first spring if its spring constant is
A)200 N/m.
B)400 N/m.
C)800 N/m.
D)1 600 N/m.
E)3 200 N/m.

Accepted Answer

The answer of A spring with spring constant k =...

Question 11

Two masses,M_A and M_B,with M_B = 2M_A,are released at the same time and allowed to fall straight down.Neglect air resistance.When we compare their kinetic energies after they have fallen for equal times,we find that A)K_B = K_A. B)K_B = 2K_A. C)K_B = 4K_A. D)K_A = 2K_B. E)K_A = 4K_B.

Accepted Answer

The answer of Two masses,M_A and M_B,with M_B = 2M_A,are...

Question 12

A spring (k = 600 N/m)is placed in a vertical position with its lower end supported by a horizontal surface.A 2.0-kg block that is initially 0.40 m above the upper end of the spring is dropped from rest onto the spring.What is the kinetic energy of the block at the instant it has fallen 0.50 m (compressing the spring 0.10 m)?
A)5.3 J
B)6.8 J
C)6.3 J
D)5.8 J
E)6.5 J

Accepted Answer

The kinetic energy of the block can be found by calculating the difference between the initial potential energy and the sum of the spring potential energy and gravitational potential energy at the point of 0.50 m descent. Initially, the block has gravitational potential energy $U_g = mgh = 2.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.50 \, \text{m} = 9.8 \, \text{J}$. The spring potential energy when compressed by 0.10 m is $U_s = \frac{1}{2}kx^2 = \frac{1}{2} \times 600 \, \text{N/m} \times (0.10 \, \text{m})^2 = 3.0 \, \text{J}$. The gravitational potential energy at the new height (0.10 m lower than the original position due to spring compression) is $U_g' = mgh' = 2.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.10 \, \text{m} = 1.96 \, \text{J}$. The kinetic energy is then $KE = U_g - (U_s + U_g') = 9.8 \, \text{J} - (3.0 \, \text{J} + 1.96 \, \text{J}) = 4.84 \, \text{J}$. However, given the options provided do not match this calculation, it appears there was a mistake in the calculation process or interpretation of the options. The correct approach involves considering the total mechanical energy conservation, but without the exact steps leading to the options provided, there's a discrepancy in the calculation presented here. The correct approach should involve calculating the initial potential energy and subtracting the energy stored in the spring and the remaining gravitational potential energy to find the kinetic energy at the point of interest.

Question 13

A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m)as shown.The speed of the mass at point A is 3.2 m/s,and at point B,it is 6.0 m/s.What is the change in mechanical energy done on the system between A and B by the force of friction? 
A)-8.9 J
B)-7.3 J
C)-8.1 J
D)-6.6 J
E)-24 J

Accepted Answer

The answer of A 1.2-kg mass is projected down a...

Question 14

A spring with spring constant k = 800 N/m is compressed 12 cm from its equilibrium position.A spring with spring constant k = 400 N/m has the same elastic potential energy as the first spring when its extension is
A)0.060 m.
B)0.085 m.
C)0.12 m.
D)0.17 m.
E)0.24 m.

Accepted Answer

The elastic potential energy of a spring is given by U = (1/2)kx^2, where k is the spring constant and x is the displacement from equilibrium position.

For the first spring with k = 800 N/m and x = 0.12 m, the elastic potential energy is: 
U1 = (1/2)(800 N/m)(0.12 m)^2 = 5.76 J

To find the extension of the second spring with k = 400 N/m that has the same elastic potential energy, we can use the fact that the elastic potential energy is the same when:

U2 = U1 
(1/2)k2x2^2 = (1/2)k1x1^2

where the subscripts 1 and 2 refer to the first and second springs, respectively.

Rearranging for x2, we get: 
x2 = sqrt(k1/k2) * x1 
x2 = sqrt(800 N/m / 400 N/m) * 0.12 m 
x2 = 0.17 m

Therefore, the answer is D) 0.17 m.

Question 15

A 0.75-kg sphere is released from rest and is moving 5.0 m/s after falling 2.0 m in a viscous medium.What is the change in mechanical energy caused by the force the viscous medium exerts on the sphere during this descent?
A)-6.1 J
B)-4.6 J
C)-5.3 J
D)-6.8 J
E)-2.7 J

Accepted Answer

The change in mechanical energy is equal to the work done by the viscous force on the sphere during its descent. This work is equal to the force multiplied by the distance over which it acts. Since the speed of the sphere is constant, the net force acting on the sphere is zero. Therefore, the viscous force must be equal in magnitude and opposite in direction to the force due to gravity.

Using the work-energy principle, we have:

Change in mechanical energy = Work done by viscous force
= Force due to viscous drag x distance
= Force due to gravity x distance  (since the two forces are equal in magnitude and opposite in direction)
= mgh
= (0.75 kg)(9.81 m/s^2)(2.0 m)
= 14.72 J

Therefore, the change in mechanical energy is -14.72 J (negative because the work done by the viscous force is negative). The closest answer choice is C, which is 5.3 J off from the correct answer.

Question 16

A 12-kg projectile is launched with an initial vertical speed of 20 m/s.It rises to a maximum height of 18 m above the launch point.What is the change in mechanical energy caused by the dissipative (air)resistive force on the projectile during this ascent?
A)-0.64 kJ
B)-0.40 kJ
C)-0.52 kJ
D)-0.28 kJ
E)-0.76 kJ

Accepted Answer

The answer of A 12-kg projectile is launched with an...

Question 17

A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m)which has its other end fixed.The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium.What is the coefficient of kinetic friction between the block and the horizontal surface?
A)0.32
B)0.45
C)0.58
D)0.19
E)0.26

Accepted Answer

We can use conservation of energy to solve this problem. The initial kinetic energy of the block equals the energy stored in the spring when the block comes to a stop:

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

where $v_0$ is the initial speed and $x$ is the displacement from equilibrium. Plugging in the numbers, we get:

$\frac{1}{2}(2.0)(2.6)^2 = \frac{1}{2}(250)(0.20)^2$

Simplifying, we get:

$6.76 = 5.0$

This is a contradiction, so we know there must be some friction present.

When the block first comes to rest at 0.20 m from equilibrium, the force provided by the spring balances the force of friction:

$kx = \mu_k mg$

where $\mu_k$ is the coefficient of kinetic friction. Plugging in the numbers, we get:

$(250)(0.20) = \mu_k (2.0)(9.81)$

Simplifying, we get:

$\mu_k = 0.45$

Therefore, the answer is choice B.

Question 18

In a given frictionless displacement of a particle,its kinetic energy increases by 25 J while its potential energy decreases by 10 J.Determine the work of the nonconservative forces acting on the particle during this displacement.
A)-15 J
B)+35 J
C)+15 J
D)-35 J
E)+55 J

Accepted Answer

According to the conservation of mechanical energy, the sum of kinetic energy and potential energy should remain constant in the absence of external forces. However, in this case, we see a change in both kinetic and potential energy, indicating the presence of nonconservative forces.

The work done by nonconservative forces is given by the difference between the initial and final mechanical energies:

Work of nonconservative forces = (Initial mechanical energy) - (Final mechanical energy)
                            = (Kinetic energy + Potential energy)_initial - (Kinetic energy + Potential energy)_final
                            = (25 J + (-10 J)) - (0 J + 0 J)
                            = 15 J

Therefore, the correct answer is C) +15 J.

Question 19

A particle is acted upon by only two forces,one conservative and one nonconservative and neither being a force of friction,as it moves from point A to point B.The kinetic energies of the particle at points A and B are equal if
A)the sum of the works of the two forces is zero.
B)the work of the conservative force is equal to the work of the nonconservative force.
C)the work of the conservative force is zero.
D)the work of the nonconservative force is zero.
E)None of the above.

Accepted Answer

When the sum of the works of the two forces (conservative and nonconservative) is zero, the total mechanical energy of the system remains constant, implying that the kinetic energy at points A and B would be equal, assuming no other energy transformations.

Question 20

Identical masses m are attached to identical springs of spring constant k suspended from the ceiling.With both masses hanging in their equilibrium positions,mass A is pulled down 10 cm and released while mass B is pushed up 10 cm and released.Which statement is correct?
A)Mass A will travel a smaller distance to its highest point than mass B will travel to its lowest point.
B)Mass A will travel a greater distance to its highest point than mass B will travel to its lowest point.
C)Masses A and B will travel equal distances between their highest and lowest points.
D)More work was done on mass A by the extending force than on mass B by the compressing force.
E)The total work done on mass A by the extending force was equal to the total work done on mass B by the compressing force.

Accepted Answer

When both masses A and B have been released, they will oscillate back and forth in a simple harmonic motion. Since the masses and the springs are identical, the oscillation period for both masses will be the same. This means that the time taken by mass A to reach its highest point and return to its starting point will be equal to the time taken by mass B to reach its lowest point and return to its starting point. Since the time taken is the same, the distance traveled by mass A from its starting point to its highest point will be equal to the distance traveled by mass B from its starting point to its lowest point. Thus, option C is the correct answer.

Quiz 7: Conservation of Energy

Two masses,M_A and M_B,with M_B = 2M_A,are released at the same time and allowed to fall straight down.Neglect air resistance.When we compare their kinetic energies after they have fallen equal distances,we find that

A 2.0-kg block slides down a fixed,rough curved track.The block has a speed of 5.0 m/s after its height above a horizontal surface has decreased by 1.8 m.Assume the block starts from rest.What is the change in mechanical energy of the block caused by the force of friction during this descent?

A 1.2-kg mass is projected up a rough circular track (radius = 0.80 m) as shown.The speed of the mass at point A is 8.4 m/s,and at point B,it is 5.6 m/s.What is the change in mechanical energy between A and B caused by the force of friction?

A 10-kg object is dropped from rest.After falling a distance of 50 m,it has a speed of 26 m/s.What is the change in mechanical energy caused by the dissipative (air) resistive force on the object during this descent?

A spring with spring constant k = 800 N/m is extended 12 cm from its equilibrium position.A spring with 6.0 cm extension from equilibrium will have the same potential energy as the first spring if its spring constant is

Two masses,M_A and M_B,with M_B = 2M_A,are released at the same time and allowed to fall straight down.Neglect air resistance.When we compare their kinetic energies after they have fallen for equal times,we find that

A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m) as shown.The speed of the mass at point A is 3.2 m/s,and at point B,it is 6.0 m/s.What is the change in mechanical energy done on the system between A and B by the force of friction?

A spring with spring constant k = 800 N/m is compressed 12 cm from its equilibrium position.A spring with spring constant k = 400 N/m has the same elastic potential energy as the first spring when its extension is

A 0.75-kg sphere is released from rest and is moving 5.0 m/s after falling 2.0 m in a viscous medium.What is the change in mechanical energy caused by the force the viscous medium exerts on the sphere during this descent?

A 12-kg projectile is launched with an initial vertical speed of 20 m/s.It rises to a maximum height of 18 m above the launch point.What is the change in mechanical energy caused by the dissipative (air) resistive force on the projectile during this ascent?

In a given frictionless displacement of a particle,its kinetic energy increases by 25 J while its potential energy decreases by 10 J.Determine the work of the nonconservative forces acting on the particle during this displacement.

A particle is acted upon by only two forces,one conservative and one nonconservative and neither being a force of friction,as it moves from point A to point B.The kinetic energies of the particle at points A and B are equal if

Identical masses m are attached to identical springs of spring constant k suspended from the ceiling.With both masses hanging in their equilibrium positions,mass A is pulled down 10 cm and released while mass B is pushed up 10 cm and released.Which statement is correct?