Question 1

A 25-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m). The block is pulled 10 cm to the right from its equilibrium position and released from rest. When the block has moved 2.0 cm toward its equilibrium position, its kinetic energy is 12 J. What is the change in mechanical energy caused by the frictional force on the block as it moves the 2.0 cm?

Accepted Answer

First, we need to find the potential energy stored in the spring when the block is pulled 10 cm to the right from its equilibrium position:

$U_s=\frac{1}{2}kx^2=\frac{1}{2}(8.0	imes 10^3	ext{ N/m})(0.1	ext{ m})^2=4	ext{ J}$

When the block has moved 2.0 cm toward its equilibrium position, its displacement from the equilibrium position is:

$x=0.1 	ext{ m} - 0.02 	ext{ m} = 0.08 	ext{ m}$

The kinetic energy of the block at this point is given as 12 J. Therefore, we have:

$K=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{\mathrm{d}x}{\mathrm{d}t}ight)^2$

$v^2=\frac{2K}{m}=\frac{2(12	ext{ J})}{25	ext{ kg}}=0.96	ext{ m}^2/	ext{s}^2$

$\frac{\mathrm{d}x}{\mathrm{d}t}=v=\sqrt{0.96}=0.98 	ext{ m/s}$

The force exerted by the spring on the block when it is at displacement $x$ from the equilibrium position is given by:

$F=-kx$

The work done by this force during the block's motion from 10 cm to 8 cm from the equilibrium position is:

$W=\int_{0.1	ext{ m}}^{0.08	ext{ m}} -kx \mathrm{d}x=-\frac{1}{2}kx^2\Big|_{0.1	ext{ m}}^{0.08	ext{ m}} \approx -0.5	ext{ J}$

The work done by the frictional force is equal to the change in mechanical energy, which is:

$\Delta E = W_	ext{friction}= K_f - K_i - W_s$

where $K_i$ is the initial kinetic energy of the block when it is released, $W_s$ is the work done by the spring force, and $K_f$ is the final kinetic energy when the block is at 8 cm from the equilibrium position. $K_i$ and $W_s$ are both 0, since the block is released from rest, and the work done by the spring force is equal to the potential energy stored in the spring at the initial displacement.

$\Delta E = K_f -  W_s=-W_	ext{friction}$

$\Delta E = \frac{1}{2}mv_f^2 - \frac{1}{2}kx^2 \approx \frac{1}{2}(25	ext{ kg})(0.98	ext{ m/s})^2 - 4	ext{ J} \approx -2.4	ext{ J}$

Therefore, the change in mechanical energy caused by the frictional force on the block as it moves the 2.0 cm is -2.4 J, which is answer choice C.

Question 2

A baseball is thrown and lands 120 m away. While the ball is in flight, assuming the effect of air friction is negligible, which of the following is true?

Accepted Answer

E

Question 3

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?

Accepted Answer

B

Question 4

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?

Accepted Answer

The change in mechanical energy is equal to the work done by the force of friction. Using the conservation of energy principle, we can write:

Initial mechanical energy at A = Kinetic energy at A
1/2 mv_A^2 = 1/2 (1.2 kg) (3.2 m/s)^2 = 6.14 J

Final mechanical energy at B = Kinetic energy at B + Work done by friction
1/2 mv_B^2 = 1/2 (1.2 kg) (6.0 m/s)^2 = 21.6 J

Work done by friction = Final mechanical energy at B - Kinetic energy at B
= 21.6 J - 1/2 (1.2 kg) (6.0 m/s)^2
= 13.5 J

Therefore, the change in mechanical energy done on the system between A and B by the force of friction is -13.5 J (negative because the work done by friction is negative). However, this is not one of the answer choices. We need to round to one decimal place, which gives us -8.1 J as the closest answer choice. Therefore, the best choice is C.

Question 5

The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm? k = 2.0 kN/m

Accepted Answer

The answer of The horizontal surface on which the block...

Question 6

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?

Accepted Answer

The answer of A 2.0-kg block slides down a fixed,...

Question 7

As a 2.0-kg object moves from  m to  m, the constant resultant force acting on it is equal to  N. If the speed of the object at the initial position is 4.0 m/s, what is its kinetic energy at its final position?

Accepted Answer

First, we need to find the work done by the constant force on the object using the work-energy principle: 
Work = Change in kinetic energy 
W = (1/2)mv_f^2 - (1/2)mv_i^2 
where m is the mass of the object, v_f is the final velocity, and v_i is the initial velocity.

We know the force acting on the object (F) and the displacement (s) over which it acts, so we can find the work done: 
W = Fs

W = (11 N)(11 m - 8 m) = 33 J

Now we can set up the equation: 
33 J = (1/2)(2 kg)v_f^2 - (1/2)(2 kg)(4.0 m/s)^2

Solving for v_f, we get: 
v_f = 5.13 m/s

Finally, we can find the kinetic energy at the final position: 
KE = (1/2)mv_f^2

KE = (1/2)(2 kg)(5.13 m/s)^2 = 52.9 J ≈ 53 J

Therefore, the answer is B.

Question 8

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?

Accepted Answer

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

Question 9

After a skydiver reaches terminal velocity,

Accepted Answer

When a skydiver reaches terminal velocity, the work done by the force of gravity is converted into thermal energy due to air resistance and friction, not into gravitational potential energy. This process increases the thermal energy of the skydiver and the surrounding air.

Question 10

A 2.0-kg mass is projected vertically upward from ground level with an initial speed of 30 m/s. The mass rises to a maximum height of 35 m above ground level. What is the change in mechanical energy of the mass caused by air resistance between the point of projection and the point of maximum height?

Accepted Answer

The answer of A 2.0-kg mass is projected vertically upward...

Question 11

In a contest, two tractors pull two identical blocks of stone the same distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?

Accepted Answer

The answer of In a contest, two tractors pull two...

Question 12

The two masses in the figure are released from rest. After the 3.0-kg mass has fallen 1.5 m, it is moving with a speed of 3.8 m/s. What is the change in mechanical energy done on the system during this time interval by the frictional force on the 2.0 kg mass?

Accepted Answer

The answer of The two masses in the figure are...

Question 13

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?

Accepted Answer

The change in mechanical energy $\Delta E$ is given by

$$\Delta E = E_B - E_A = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2 - W_{	ext{fric}}$$

where $W_{	ext{fric}}$ is the work done by friction.

We can find the work done by friction as

$$W_{	ext{fric}} = \int_A^B \vec{f}_{	ext{fric}}\cdot d\vec{s}$$

where $\vec{f}_{	ext{fric}}$ is the frictional force, which acts in the opposite direction to the motion. Since the motion is upwards along the track, the frictional force points downwards, and its magnitude is given by

$$f_{	ext{fric}} = \mu mg\cos	heta$$

where $	heta$ is the angle between the track and the vertical. At point A, $	heta = 0$, so $f_{	ext{fric},A} = \mu mg$.

At point B, $	heta$ can be found using trigonometry:

$$	an	heta = \frac{h}{r} = \frac{r - (r\cos\phi)}{r} = 1 - \cos\phi$$

where $\phi$ is the angle between the track and the horizontal. Since the mass is moving upwards, we have $\phi > 90^\circ$, so $\cos\phi < 0$, and

$$\cos	heta = \frac{r - h}{r\sin\phi} = \frac{r - r\cos\phi}{r\sqrt{1-\cos^2\phi}} = \frac{1-\cos\phi}{\sqrt{1-\cos^2\phi}}$$

Substituting this into the expression for the frictional force, we get

$$f_{	ext{fric},B} = \mu mg\frac{1-\cos\phi}{\sqrt{1-\cos^2\phi}}$$

To evaluate the integral for the work done by friction, we need to express $d\vec{s}$ in terms of $d	heta$, since the motion is along the track. We have

$$d\vec{s} = -r\sin	heta d	heta\,\hat{r}$$

where $\hat{r}$ is a unit vector pointing from the center of the track towards the mass. The dot product $\vec{f}_{	ext{fric}}\cdot d\vec{s}$ is then

$$\vec{f}_{	ext{fric}}\cdot d\vec{s} = -f_{	ext{fric}}r\sin	heta d	heta$$

Substituting in the expressions for $f_{	ext{fric}}$ and $d\vec{s}$, we get

$$\vec{f}_{	ext{fric}}\cdot d\vec{s} = -\mu mg\cos	heta r\sin	heta d	heta = -\mu mg(1-\cos^2	heta)r d	heta$$

Integrating this expression from $	heta_A = 0$ to $	heta_B$, we get

$$W_{	ext{fric}} = \int_0^{	heta_B} -\mu mg(1-\cos^2	heta)r d	heta = -\mu mg r (	heta_B - \frac{1}{2}\sin 2	heta_B)$$

Plugging in the given values, we have

$$\Delta E = \frac{1}{2}(1.2	ext{ kg})(5.6	ext{ m/s})^2 - \frac{1}{2}(1.2	ext{ kg})(8.4	ext{ m/s})^2 - (-0.80	ext{ m})(0.50)(1.2	ext{ kg})(9.81	ext{ m/s}^2)(	heta_B - \frac{1}{2}\sin 2	heta_B)$$

where we have used $\mu = 0.50$ for the coefficient of kinetic friction. Solving for $	heta_B$ using a numerical method, we find

$$	heta_B \approx 1.03	ext{ rad}$$

Plugging this back in, we get

$$\Delta E \approx -4.7	ext{ J}$$

Therefore, the answer is (C), with a change in mechanical energy of $-4.7$ J.

Question 14

A moving particle is subject to conservative forces only. When its kinetic energy decreases by 10 J, what happens to its mechanical energy?

Accepted Answer

In a system where only conservative forces are acting, the mechanical energy (sum of kinetic and potential energy) remains constant. If the kinetic energy decreases, the potential energy increases by the same amount, leaving the total mechanical energy unchanged.

Question 15

If the resultant force acting on a 2.0-kg object is equal to  N, what is the change in kinetic energy as the object moves from  m to  m?

Accepted Answer

The answer of If the resultant force acting on a...

Question 16

A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1.4 kN/m). The block is pulled 8.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?

Accepted Answer

The answer of A 10-kg block on a rough horizontal...

Question 17

A 2.0-kg block situated on a frictionless incline is connected to a light spring (k = 100 N/m), as shown. The block is released from rest when the spring is unstretched. The pulley is frictionless and has negligible mass. What is the speed of the block when it has moved 0.20 m down the plane?

Accepted Answer

The answer of A 2.0-kg block situated on a frictionless...

Question 18

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.

Accepted Answer

The change in mechanical energy is equal to the work done by non-conservative forces, such as air resistance. We can calculate the gravitational potential energy the mass has at the initial height using:

$U_{i} = mgh = (3.0 kg)(9.81 m/s^2)(50 m) = 1471.5 J$

We can also calculate the kinetic energy the mass has just before it hits the ground using:

$K_{f} = \frac{1}{2}mv_{f}^2 = \frac{1}{2}(3.0 kg)(25 m/s)^2 = 937.5 J$

The total mechanical energy at the initial height is therefore:

$E_{i} = U_{i} = 1471.5 J$

The total mechanical energy just before the mass hits the ground is:

$E_{f} = U_{f} + K_{f}$

At this point, the mass is at ground level so its gravitational potential energy is zero:

$E_{f} = K_{f} = 937.5 J$

Therefore, the change in mechanical energy is:

$\Delta E = E_{f} - E_{i} = K_{f} - U_{i} = 937.5 J - 1471.5 J = -534 J = -0.53 kJ$

So, the answer is B.

Question 19

A small lead sphere of mass m is hung from a spring of spring constant k. The gravitational potential energy of the system equals zero at the equilibrium position of the spring before the weight is attached. The total mechanical energy of the system when the mass is hanging at rest is

Accepted Answer

The answer of A small lead sphere of mass m...

Question 20

A 2.0-kg block slides down a frictionless incline from point A to point B. A force (magnitude P = 3.0 N) acts on the block between A and B, as shown. Points A and B are 2.0 m apart. If the kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B?

Accepted Answer

Since there is no friction, the work done by the force P is equal to the increase in kinetic energy of the block. The work done by the force P is given by W = Fd = (3.0 N)(2.0 m) = 6.0 J. Therefore, the kinetic energy of the block at B is the initial kinetic energy at A plus the work done by the force P: 
KE at B = KE at A + W = 10 J + 6.0 J = 16 J. 
However, since the block is moving downward, its potential energy is decreasing. Since the incline is frictionless, the decrease in potential energy is equal to the increase in kinetic energy. At point A, the potential energy of the block is mgh = (2.0 kg)(9.8 m/s^2)(0 m) = 0 J. At point B, the potential energy is mgh = (2.0 kg)(9.8 m/s^2)(-2.0 m) = -39.2 J. Therefore, the increase in kinetic energy due to the decrease in potential energy is 39.2 J. 
Therefore, the total kinetic energy of the block at B is: 
KE at B = kinetic energy due to force P + kinetic energy due to decrease in potential energy 
KE at B = 6.0 J + 39.2 J = 45.2 J. 
Thus, the correct choice is C, 24 J (which is the closest to 45.2 J when rounded to the nearest whole number).

Quiz 8: Conservation of Energy