Question 1

A $20.0 \mathrm{~N}$ force is applied at an angle of 40.0 degrees above the horizontal to a $4.00 \mathrm{~kg}$ box. The box moves a horizontal distance of 4.00 meters. Friction is negligible. The work done by the $20.0 \mathrm{~N}$ force is

A)

46.3 \mathrm{~J}

.
B)

50.1 \mathrm{~J}

.
C)

75.0 \mathrm{~J}

.
D)

40.5 \mathrm{~J}

.
E)

61.3 \mathrm{~J}

.

61.3 \mathrm{~J}

.

Accepted Answer

The work done by the force is calculated using the formula $W = Fd\cos(\theta)$, where $W$ is the work done, $F$ is the magnitude of the force, $d$ is the distance moved by the object, and $\theta$ is the angle of the force with respect to the direction of motion. Here, $F = 20.0 \, \mathrm{N}$, $d = 4.00 \, \mathrm{m}$, and $\theta = 40.0^\circ$. Thus, $W = 20.0 \times 4.00 \times \cos(40.0^\circ) \approx 61.3 \, \mathrm{J}$.

Question 2

A $6.00 \mathrm{~kg}$ box is pulled up an $8.00 \mathrm{~m}$ long incline (with friction) at an angle of 30.0 degrees by a force of $50.0 \mathrm{~N}$ parallel to the incline. The coefficient of kinetic friction is 0.100 . The work done against friction is

A)

40.7 \mathrm{~J}

.
B)

30.3 \mathrm{~J}

.
C)

22.0 \mathrm{~J}

.
D)

26.5 \mathrm{~J}

.
E)

35.2 \mathrm{~J}

40.7 \mathrm{~J}

.

Accepted Answer

The work done against friction is calculated using the formula $W = f_k \cdot d$, where $f_k$ is the force of kinetic friction and $d$ is the distance. The force of kinetic friction is given by $f_k = \mu_k \cdot N$, where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force. The normal force for an incline is calculated as $N = mg \cos(\theta)$, where $m$ is the mass, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $\theta$ is the angle of the incline. For this problem:- $m = 6.00 \, \mathrm{kg}$- $\mu_k = 0.100$- $\theta = 30.0$ degrees- $d = 8.00 \, \mathrm{m}$- $g = 9.8 \, \mathrm{m/s^2}$First, calculate the normal force:$$N = mg \cos(\theta) = 6.00 \times 9.8 \times \cos(30.0^\circ) \approx 50.9 \, \mathrm{N}$$Then, calculate the force of kinetic friction:$$f_k = \mu_k \cdot N = 0.100 \times 50.9 \approx 5.09 \, \mathrm{N}$$Finally, calculate the work done against friction:$$W = f_k \cdot d = 5.09 \times 8.00 \approx 40.7 \, \mathrm{J}$$Therefore, the correct answer is $40.7 \, \mathrm{J}$, which is option A.

Question 3

The graph shows the force on an object as it moves a distance $x$. What is the work done by the force as the object moves from $2.0 \mathrm{~m}$ to $8.0 \mathrm{~m}$ ?&#10; &#10;A) $52 \mathrm{~J}$&#10;B) $50 \mathrm{~J}$&#10;C) $35 \mathrm{~J}$&#10;D) $60 \mathrm{~J}$&#10;E) $48 \mathrm{~J}$

Accepted Answer

$48 \mathrm{~J}$&#10;

Question 4

The graph shows the force on an object as it moves a distance $\mathrm{x}$. What is the work done by the force as the object moves from $0.0 \mathrm{~m}$ to $6.0 \mathrm{~m}$ ?&#10; &#10;A) $8.0 \mathrm{~J}$&#10;B) $14 \mathrm{~J}$&#10;C) $12 \mathrm{~J}$&#10;D) $6.0 \mathrm{~J}$&#10;E) $10 \mathrm{~J}$

Accepted Answer

The answer of  The graph shows the force on...

Question 5

A 5.00 gram bullet has a velocity of $300 \mathrm{~m} / \mathrm{s}$ . It strikes a block of wood and penetrates to a depth of 2.00 $\mathrm{cm}$ . The average force of the block of wood on the bullet is

A)

21,400 \mathrm{~N}

.
B)

11,250 \mathrm{~N}

.
C)

15,200 \mathrm{~N}

.
D)

9,340 \mathrm{~N}

.
E)

19,500 \mathrm{~N}

.

Accepted Answer

To find the average force, we can use the work-energy principle. The work done by the force is equal to the change in kinetic energy of the bullet. The kinetic energy (KE) of the bullet before it stops is given by $KE = \frac{1}{2}mv^2$, where $m = 0.005 \, \text{kg}$ (since 1 gram = 0.001 kg) and $v = 300 \, \text{m/s}$. The change in kinetic energy is equal to the work done by the force, $W = Fd$, where $F$ is the average force and $d = 0.02 \, \text{m}$ (since 1 cm = 0.01 m). Solving for $F$, we get $F = \frac{KE}{d}$. Substituting the values, we find $F = \frac{0.5 \times 0.005 \times 300^2}{0.02} = 11,250 \, \text{N}$.

Question 6

A $2,000 \mathrm{~kg}$ car traveling at $30.0 \mathrm{mi} / \mathrm{h}$ skids to a stop in 60.0 meters. The force of friction during the skid is $(1.0 \mathrm{mi} / \mathrm{h}=0.447 \mathrm{~m} / \mathrm{s})$

A)

2,400 \mathrm{~N}

.
B)

3,000 \mathrm{~N}

.
C)

4,200 \mathrm{~N}

.
D)

3,750 \mathrm{~N}

E)

4,500 \mathrm{~N}

Accepted Answer

The answer of A $2,000 \mathrm{~kg}$ car traveling at \(30.0...

Question 7

The explosion in a cannon exerts an average force of $30,000 \mathrm{~N}$ for $\mathrm{L}$ meters, the length of the cannon. What length of the cannon would be necessary to shoot a $2.0 \mathrm{~kg}$ projectile from the surface of the Earth to a distance of $6.84 \times 10^{8} \mathrm{~m}$ from the center of the Earth (the same as the distance to the Moon)? $(\mathrm{G}=6.67 \times$ $10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg} 2, \mathrm{ME}=5.97 \times 10^{24} \mathrm{~kg}$ , and $\mathrm{RE}=6.37 \times 10^{6} \mathrm{~m}$ .)

A)

3.9 \mathrm{~km}

.
B)

4.1 \mathrm{~km}

.
C)

5.0 \mathrm{~km}

.
D)

3.0 \mathrm{~km}

.
E)

2.7 \mathrm{~km}

.

Accepted Answer

To find the necessary length of the cannon, we first calculate the work done by the cannon to provide the projectile with enough energy to reach the specified distance from the center of the Earth. This energy must be equal to the change in gravitational potential energy between the surface of the Earth and the distance to the Moon.The gravitational potential energy at a distance $r$ from the center of the Earth is given by $U = -\frac{GM_Em}{r}$, where $G$ is the gravitational constant, $M_E$ is the mass of the Earth, and $m$ is the mass of the projectile.The change in potential energy ($\Delta U$) as the projectile moves from the surface of the Earth ($r = R_E$) to the distance of the Moon ($r = 6.84 \times 10^8 \, m$) is:$$\Delta U = U_{final} - U_{initial} = -\frac{GM_Em}{6.84 \times 10^8} - \left(-\frac{GM_Em}{R_E}\right)$$Substituting the given values:$$\Delta U = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(2)}{6.84 \times 10^8} + \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(2)}{6.37 \times 10^6}$$Solving this gives $\Delta U \approx 6.25 \times 10^{10} \, J$.The work done by the cannon is equal to the force exerted by the explosion times the distance over which it acts ($W = F \cdot L$), and this work must equal the change in potential energy to achieve the desired distance. Therefore, $30,000 \cdot L = 6.25 \times 10^{10}$, solving for $L$ gives $L \approx 2.08 \times 10^6 \, m$ or $2080 \, km$.However, the calculation above leads to a misunderstanding in the interpretation of the physical scenario or a miscalculation. The correct approach involves equating the work done by the cannon to the gravitational potential energy difference needed to escape Earth's gravity and reach the Moon's distance, but the calculation provided does not accurately reflect the correct answer options given, indicating a mistake in the calculation process. The correct method involves calculating the necessary kinetic energy to reach the escape velocity or the specific energy required to reach the Moon's orbit and then finding the cannon length that would provide this energy through work done by the force of the explosion. The discrepancy suggests a need to reevaluate the calculation steps or assumptions made.

Question 8

A$20.0 \mathrm{~kg}$ box slides down a $12.0 \mathrm{~m}$ long incline at an angle of 30.0 degrees with the horizontal. A force of $50.0 \mathrm{~N}$ is applied to the box to try to prevent it from sliding down the incline. The applied force makes an angle of 10.00 degrees to the incline and above the incline. If the incline has a coefficient of kinetic friction of 0.100 , then the increase in the kinetic energy of the box is&#10;A) $392 \mathrm{~J}$.&#10;B) $300 \mathrm{~J}$.&#10;C) $275 \mathrm{~J}$.&#10;D) $203 \mathrm{~J}$.&#10;E) $425 \mathrm{~J}$.

Accepted Answer

The increase in kinetic energy of the box can be calculated by finding the net work done on the box. The net work done is the sum of the work done by gravity, the work done by the applied force, and the work done by friction.1. Work done by gravity ($W_g$):$$W_g = mgh = mgd\sin(\theta) = (20.0\,\text{kg})(9.8\,\text{m/s}^2)(12.0\,\text{m})\sin(30^\circ) = 1176\,\text{J}$$2. Work done by the applied force ($W_f$):The component of the applied force along the incline is $F\cos(\phi)$, where $F = 50.0\,\text{N}$ and $\phi = 10^\circ$.$$W_f = Fd\cos(\phi) = (50.0\,\text{N})(12.0\,\text{m})\cos(10^\circ) \approx 589.4\,\text{J}$$However, since the force is applied to prevent the box from sliding down, it does negative work (it acts opposite to the direction of motion), so we need to consider its value as negative:$$W_f = -589.4\,\text{J}$$3. Work done by friction ($W_{fr}$):The normal force ($N$) is the sum of the component of the gravitational force perpendicular to the incline and the component of the applied force perpendicular to the incline:$$N = mg\cos(\theta) + F\sin(\phi) = (20.0\,\text{kg})(9.8\,\text{m/s}^2)\cos(30^\circ) + (50.0\,\text{N})\sin(10^\circ)$$$$N \approx 196\,\text{N} + 8.7\,\text{N} = 204.7\,\text{N}$$The frictional force ($f_k$) is $f_k = \mu_k N$, where $\mu_k = 0.100$.$$f_k = 0.100 \times 204.7\,\text{N} = 20.47\,\text{N}$$The work done by friction is:$$W_{fr} = -f_k d = -(20.47\,\text{N})(12.0\,\text{m}) = -245.64\,\text{J}$$4. Total work done ($W_{total}$) is the sum of all works:$$W_{total} = W_g + W_f + W_{fr} = 1176\,\text{J} - 589.4\,\text{J} - 245.64\,\text{J} \approx 341\,\text{J}$$It seems there was a mistake in my calculations for the work done by the applied force and its interpretation, leading to an incorrect total work and thus an incorrect choice. The correct approach should focus on accurately calculating the net force and the work done by all forces involved, including correctly applying the direction of the applied force and friction. Given the options, none of my calculations directly match, indicating a need to reassess the calculations and the principles applied. The correct answer should reflect the net work done on the box, considering gravitational force, applied force, and friction correctly.

Question 9

A $20.0 \mathrm{~kg}$ box slides down a $12.0 \mathrm{~m}$ long incline at an angle of 30.0 degrees with the horizontal. A force of $50.0 \mathrm{~N}$ is applied to the box to try to prevent it from sliding down the incline. The applied force makes an angle of 10.00 degrees to the incline and above the incline. If the incline has a coefficient of kinetic friction of 0.100 , then the increase in the kinetic energy of the box is

A)

392 \mathrm{~J}

.
B)

300 \mathrm{~J}

.
C)

275 \mathrm{~J}

.
D)

203 \mathrm{~J}

.
E)

425 \mathrm{~J}

.

Accepted Answer

The answer of A $20.0 \mathrm{~kg}$ box slides down a...

Question 10

A $20.0 \mathrm{~kg}$ box slides down a $12.0 \mathrm{~m}$ long incline at an angle of 3.0 degrees with the horizontal. A force of $50.0 \mathrm{~N}$ is applied to the box to try to prevent it from sliding down the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100 , then the increase in the kinetic energy of the box is

A)

525 \mathrm{~J}

.
B)

455 \mathrm{~J}

.
C)

300 \mathrm{~J}

.
D)

372 \mathrm{~J}

.
E)

410 \mathrm{~J}

.

Accepted Answer

The increase in kinetic energy can be calculated by finding the work done on the box. The work done by gravity is $W_g = mgh$, where $h = 12\sin(30^\circ)$ is the vertical height of the incline. The work done by the applied force is $W_f = Fd\cos(\theta)$, where $F = 50\,N$, $d = 12\,m$, and $\theta = 10^\circ$ is the angle between the force and the direction of motion. The total work done is the sum of these, which equals the increase in kinetic energy. Given $m = 20.0\,kg$, $g = 9.8\,m/s^2$, the calculation yields an increase in kinetic energy of $585\,J$.

Question 11

A $2500 \mathrm{~kg}$ car accelerates from rest to a velocity of $30.0 \mathrm{~m} / \mathrm{s}$ in 8.00 seconds. The work done by the engine to accelerate the car is&#10;A) $1.13 \times 106 \mathrm{~J}$.&#10;B) $2.01 \times 106 \mathrm{~J}$.&#10;C) $1.85 \times 106 \mathrm{~J}$.&#10;D) $3.10 \times 106 \mathrm{~J}$.&#10;E) $2.56 \times 106 \mathrm{~J}$.

Accepted Answer

The answer of A $2500 \mathrm{~kg}$ car accelerates from rest...

Question 12

A $3.00 \mathrm{~kg}$ pendulum bob on a string $2.00 \mathrm{~m}$ long is released with a velocity of $2.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 45.0 degrees with the vertical. What is the tension in the string at the highest point of its motion?

A)

31.6 \mathrm{~N}

B)

12.5 \mathrm{~N}

C)

28.9 \mathrm{~N}

D)

17.8 \mathrm{~N}

E)

21.4 \mathrm{~N}

Accepted Answer

The answer of A $3.00 \mathrm{~kg}$ pendulum bob on a...

Question 13

A $2.00 \mathrm{~kg}$ pendulum bob on a string $3.00 \mathrm{~m}$ long is released with a velocity of $1.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the tension in the string at the highest point of its motion?

A)

30.8 \mathrm{~N}

B)

10.5 \mathrm{~N}

C)

27.5 \mathrm{~N}

D)

16.6 \mathrm{~N}

E)

20.1 \mathrm{~N}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 14

A $3.00 \mathrm{~kg}$ pendulum bob on a string $2.00 \mathrm{~m}$ long, is released with a velocity of $2.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 45.0 degrees with the vertical. What is the tangential acceleration of the bob at the highest point of its motion?

A)

6.04 \mathrm{~m} / \mathrm{s}^{2}

B)

7.80 \mathrm{~m} / \mathrm{s}^{2}

C)

6.89 \mathrm{~m} / \mathrm{s}^{2}

D)

8.20 \mathrm{~m} / \mathrm{s}^{2}

E)

9.02 \mathrm{~m} / \mathrm{s}^{2}

Accepted Answer

The answer of A $3.00 \mathrm{~kg}$ pendulum bob on a...

Question 15

A $2.00 \mathrm{~kg}$ pendulum bob on a string $2.00 \mathrm{~m}$ long, is released with a velocity of $1.50 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the tangential acceleration of the bob at the highest point of its motion?

A)

6.25 \mathrm{~m} / \mathrm{s}^{2}

B)

7.25 \mathrm{~m} / \mathrm{s}^{2}

C)

5.33 \mathrm{~m} / \mathrm{s}^{2}

D)

7.00 \mathrm{~m} / \mathrm{s}^{2}

E)

5.77 \mathrm{~m} / \mathrm{s}^{2}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 16

A $2.00 \mathrm{~kg}$ pendulum bob on a string $1.50 \mathrm{~m}$ long, is released with a velocity of $2.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the speed of the bob at the bottom of the swing?

A)

4.32 \mathrm{~m} / \mathrm{s}

B)

3.75 \mathrm{~m} / \mathrm{s}

C)

3.04 \mathrm{~m} / \mathrm{s}

D)

4.00 \mathrm{~m} / \mathrm{s}

E)

2.82 \mathrm{~m} / \mathrm{s}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 17

A $2.00 \mathrm{~kg}$ pendulum bob on a string $3.00 \mathrm{~m}$ long is released with a velocity of $1.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the tension in the string at the highest point of its motion?

A)

30.8 \mathrm{~N}

B)

10.5 \mathrm{~N}

C)

27.5 \mathrm{~N}

D)

16.6 \mathrm{~N}

E)

20.1 \mathrm{~N}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 18

A $2.00 \mathrm{~kg}$ pendulum bob on a string $1.50 \mathrm{~m}$ long, is released with a velocity of $2.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the speed of the bob at the bottom of the swing?

A)

4.32 \mathrm{~m} / \mathrm{s}

B)

3.75 \mathrm{~m} / \mathrm{s}

C)

3.04 \mathrm{~m} / \mathrm{s}

D)

4.00 \mathrm{~m} / \mathrm{s}

E)

2.82 \mathrm{~m} / \mathrm{s}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 19

A $2.00 \mathrm{~kg}$ pendulum bob on a string $3.00 \mathrm{~m}$ long is released with a velocity of $1.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the tension in the string at the highest point of its motion?

A)

30.8 \mathrm{~N}

B)

10.5 \mathrm{~N}

C)

27.5 \mathrm{~N}

D)

16.6 \mathrm{~N}

E)

20.1 \mathrm{~N}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 20

A $2.00 \mathrm{~kg}$ pendulum bob on a string $3.00 \mathrm{~m}$ long is released with a velocity of $1.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the tension in the string at the highest point of its motion?

A)

30.8 \mathrm{~N}

B)

10.5 \mathrm{~N}

C)

27.5 \mathrm{~N}

D)

16.6 \mathrm{~N}

E)

20.1 \mathrm{~N}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 21

A $2.00 \mathrm{~kg}$ pendulum bob on a string $3.00 \mathrm{~m}$ long is released with a velocity of $1.00 \mathrm{~m} / \mathrm{s}$ when the support string makes an angle of 30.0 degrees with the vertical. What is the tension in the string at the highest point of its motion?

A)

30.8 \mathrm{~N}

B)

10.5 \mathrm{~N}

C)

27.5 \mathrm{~N}

D)

16.6 \mathrm{~N}

E)

20.1 \mathrm{~N}

Accepted Answer

The answer of A $2.00 \mathrm{~kg}$ pendulum bob on a...

Question 22

A $75.0 \mathrm{~kg}$ skier, starting from rest, slides down a $75.0 \mathrm{~m}$ high slope without friction. The velocity of the skier at the bottom of the slope is&#10;A) $40.5 \mathrm{~m} / \mathrm{s}$.&#10;B) $50.0 \mathrm{~m} / \mathrm{s}$.&#10;C) $20.6 \mathrm{~m} / \mathrm{s}$.&#10;D) $29.7 \mathrm{~m} / \mathrm{s}$.&#10;E) $38.3 \mathrm{~m} / \mathrm{s}$

Accepted Answer

The answer of A $75.0 \mathrm{~kg}$ skier, starting from rest,...

Question 23

A student lifts a weight of $10.0 \mathrm{~N}$ a distance of 0.500 meters. The energy needed to do this work in calories is $(1 \mathrm{cal}=4.186$ Joules)&#10;A) $1.19 \mathrm{cal}$.&#10;B) $2.30 \mathrm{cal}$.&#10;C) $0.570 \mathrm{cal}$.&#10;D) $0.750 \mathrm{cal}$.&#10;E) $1.75 \mathrm{cal}$.

Accepted Answer

The answer of A student lifts a weight of \(10.0...

Question 24

An $8000 \mathrm{~kg}$ satellite is launched from the surface of the Earth into outer space. What initial kinetic energy is needed by the satellite in order to reach a great (i.e., infinite) distance from the Earth, neglecting the effects of air resistance in the atmosphere? $\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}, \mathrm{M}_{\mathrm{E}}=5.97 \times 10^{24} \mathrm{~kg}, \mathrm{R}_{\mathrm{E}}=6.37 \times 10^{6} \mathrm{~m}\right.$ .)

A)

5.00 \times 1011 \mathrm{~J}

.
B)

3.00 \times 1011 \mathrm{~J}

.
C)

2.35 \times 1011 \mathrm{~J}

.
D)

4.03 \times 1011 \mathrm{~J}

.
E)

3.57 \times 1011 \mathrm{~J}

.

Accepted Answer

The answer of An $8000 \mathrm{~kg}$ satellite is launched from...

Question 25

An $8000 \mathrm{~kg}$ satellite is launched from the surface of the Earth and injected into a circular orbit at an altitude of $100 \mathrm{~km}$ above the surface of the Earth. The kinetic energy of the satellite in the circular orbit is $(G=6.67 \times$ $10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}, \mathrm{M}_{\mathrm{E}}=5.97 \times 1024 \mathrm{~kg}, \mathrm{R}_{\mathrm{E}}=6.37 \times 10^{6} \mathrm{~m}$ )

A)

5.02 \times 1011 \mathrm{~J}

.
B)

3.45 \times 1011 \mathrm{~J}

.
C)

2.01 \times 1011 \mathrm{~J}

.
D)

4.25 \times 1011 \mathrm{~J}

.
E)

2.46 \times 10^{11} \mathrm{~J}

.

Accepted Answer

The answer of An $8000 \mathrm{~kg}$ satellite is launched from...

Question 26

An $8000 \mathrm{~kg}$ satellite is launched from the surface of the Earth into outer space. What initial kinetic energy is needed by the satellite in order to reach a great (i.e., infinite) distance from the Earth, neglecting the effects of air resistance in the atmosphere? $\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}, \mathrm{M}_{\mathrm{E}}=5.97 \times 10^{24} \mathrm{~kg}, \mathrm{R}_{\mathrm{E}}=6.37 \times 10^{6} \mathrm{~m}\right.$ .)

A)

5.00 \times 1011 \mathrm{~J}

.
B)

3.00 \times 1011 \mathrm{~J}

.
C)

2.35 \times 1011 \mathrm{~J}

.
D)

4.03 \times 1011 \mathrm{~J}

.
E)

3.57 \times 1011 \mathrm{~J}

.

Accepted Answer

The answer of An $8000 \mathrm{~kg}$ satellite is launched from...

Question 27

The graph shows the force on an object as it moves a distance $\mathrm{x}$. What is the work done by the force as the object moves from $0.0 \mathrm{~m}$ to $10.0 \mathrm{~m}$ ?&#10; &#10;A) $52 \mathrm{~J}$&#10;B) $75 \mathrm{~J}$&#10;C) $34 \mathrm{~J}$&#10;D) $45 \mathrm{~J}$&#10;E) $64 \mathrm{~J}$

Accepted Answer

The answer of  The graph shows the force on...

Question 28

The graph shows the force on an object as it moves a distance $\mathrm{x}$. What is the work done by the force as the object moves from $0.0 \mathrm{~m}$ to $4.0 \mathrm{~m}$ ?&#10; &#10;A) $10 \mathrm{~J}$&#10;B) $6.0 \mathrm{~J}$&#10;C) $8.0 \mathrm{~J}$&#10;D) $14 \mathrm{~J}$&#10;E) $12 \mathrm{~J}$

Accepted Answer

The answer of  The graph shows the force on...

Question 29

A spring is stretched from $0.500 \mathrm{~m}$ to $0.800 \mathrm{~m}$ . Assume the unstretched position is $0.00 \mathrm{~m}$ . If the spring constant of the spring is $10.0 \mathrm{~N} / \mathrm{m}$ , what is the work done on the spring?

A)

3.01 \mathrm{~J}

B)

1.95 \mathrm{~J}

C)

2.21 \mathrm{~J}

D)

2.45 \mathrm{~J}

E)

2.75 \mathrm{~J}

Accepted Answer

The answer of A spring is stretched from $0.500 \mathrm{~m}$...

Question 30

A car has a mass of 2,000 kg and travels at $20.0 \mathrm{~m} / \mathrm{s}$ . If the drag force is $100 \mathrm{~N}$ , then the power the engines have to provide to keep moving at constant speed is

A)

3.9 \mathrm{~kW}

.
B)

2.0 \mathrm{~kW}

.
C)

4.2 \mathrm{~kW}

.
D)

2.9 \mathrm{~kW}

.
E)

3.4 \mathrm{~kW}

.

Accepted Answer

The answer of A car has a mass of 2,000...

Question 31

During a basketball game, a player shoots a ball from half-court. When the ball reaches its maximum height of $4.5 \mathrm{~m}$ above the floor, it is moving at $5.0 \mathrm{~m} / \mathrm{s}$ . If the ball was released from $2.5 \mathrm{~m}$ above the floor, what was the angle above the horizontal of the ball's initial velocity?

A)

58^{\circ}

B)

54^{\circ}

C)

39^{\circ}

D)

51^{\circ}

E)

32^{\circ}

F)

36^{\circ}

Accepted Answer

The answer of During a basketball game, a player shoots...

Question 32

An asteroid of mass $1.4 \times 10^{14} \mathrm{~kg}$ is observed to have an escape speed equal to a typical person's maximum jumping speed, $3.0 \mathrm{~m} / \mathrm{s}$. What is its radius?&#10;A) $1.0 \mathrm{~km}$&#10;B) $0.5 \mathrm{~km}$&#10;C) $2.1 \mathrm{~km}$&#10;D) $6.3 \mathrm{~km}$

Accepted Answer

The answer of An asteroid of mass \(1.4 \times 10^{14}...

Question 33

An asteroid of radius $2.1 \mathrm{~km}$ is observed to have an escape speed equal to a typical person's maximum jumping speed, $3.0 \mathrm{~m} / \mathrm{s}$. What is its mass?&#10;A) $2.8 \times 1014 \mathrm{~kg}$&#10;B) $4.2 \times 1014 \mathrm{~kg}$&#10;C) $0.5 \times 1014 \mathrm{~kg}$&#10;D) $1.4 \times 1014 \mathrm{~kg}$&#10;E) $5.7 \times 10^{14} \mathrm{~kg}$

Accepted Answer

The answer of An asteroid of radius $2.1 \mathrm{~km}$ is...

Question 34

A comet is observed to pass $1.50 \times 10^{8} \mathrm{~m}$ from the surface of the Sun. The Sun's radius is $6.96 \times 10^{8}$ $\mathrm{m}$ . What is the speed of the comet at this point, if its speed when passing the Earth's orbit, $1.50 \times 10^{11} \mathrm{~m}$ from the Sun's center, is $25 \mathrm{~km} / \mathrm{s}$ ? $\left(\mathrm{M}_{\mathrm{S}}=1.989 \times 10^{30} \mathrm{~kg}\right.$ . $)$

A)

562 \mathrm{~km} / \mathrm{s}

B)

585 \mathrm{~km} / \mathrm{s}

C)

618 \mathrm{~km} / \mathrm{s}

D)

559 \mathrm{~km} / \mathrm{s}

Accepted Answer

The answer of A comet is observed to pass \(1.50...

Question 35

An $0.10 \mathrm{~g}$ flea, having leapt from the surface of a dog's cranium, is observed to be moving at $1.25 \mathrm{~m} / \mathrm{s}$ when it is $5.00 \mathrm{~cm}$ above the position from which it leapt. What was the elastic potential energy stored in its legs before its leap?

A)

4.9 \times 10^{-5} \mathrm{~J}

B)

5.7 \times 10^{-3} \mathrm{~J}

C)

1.3 \times 10^{-4} \mathrm{~J}

D)

5.7 \times 10^{-4} \mathrm{~J}

E)

1.3 \times 10^{-3} \mathrm{~J}

F)

7.8 \times 10^{-5} \mathrm{~J}

Accepted Answer

The answer of An $0.10 \mathrm{~g}$ flea, having leapt from...

Deck 6: Conservation of Energy