Question 1

A spring with a spring constant of 8.0 N/m is initially compressed to 5.0 cm.The work done as it is compressed another 15 cm,to a total compression of 20 cm,is
A) 0.85 J.
B) 0.54 J.
C) 0.25 J.
D) 0.15 J.
E) 0.09 J.

Accepted Answer

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Question 2

A spring with a spring constant of 8.0 N/m is compressed to 20 cm.If the spring is uncompressed initially,then the work done is
A) 0.12 J.
B) 0.16 J.
C) 0.19 J.
D) 0.22 J.
E) 0.26 J.

Accepted Answer

The work done on a spring is given by the formula $ W = \frac{1}{2} k x^2 $, where $ k $ is the spring constant and $ x $ is the compression distance. Here, $ k = 8.0 \, \text{N/m} $ and $ x = 0.20 \, \text{m} $. So, $ W = \frac{1}{2} \times 8.0 \times (0.20)^2 = 0.16 \, \text{J} $.

Question 3

An 8000 kg satellite is launched from the surface of the earth,and injected into a circular orbit at an altitude of 100 km above the surface of the earth.The gravitational potential energy of the satellite in the circular orbit is (G = 6.67 $\times$ 10^-11 Nm²/kg², Me = 5.98 $\times$ 10²⁴ kg, Re = 6.38 $\times$ 10⁶ m) A) 5.01 $\times$ 10¹¹ J. B) 4.92 $\times$ 10¹¹ J. C) 4.75 $\times$ 10¹¹ J. D) 4.02 $\times$ 10¹¹ J. E) 3.85 $\times$ 10¹¹ J.

Accepted Answer

The gravitational potential energy of an object in circular orbit is given by the formula: U = -GMm/2r
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit (which is the sum of the radius of the Earth and the altitude of the orbit).
Plugging in the values, we get:
U = -[(6.67e-11 Nm^2/kg^2)×(5.98e24 kg)×(8000 kg)]/[2×(6.38e6 m + 1e5 m)]
U ≈ -4.92×10^11 J
The negative sign indicates that the potential energy is bound (i.e., the satellite is gravitationally attracted to the Earth). However, we are only interested in the magnitude of the energy, which is 4.92×10^11 J (option B).

Question 4

An 8000 kg satellite is launched from the surface of the earth,and injected into a circular orbit at an altitude of 100 km above the surface of the earth.The kinetic energy of the satellite in the circular orbit is (G = 6.67 $\times$ 10^-11 Nm²/kg², Me = 5.98 $\times$ 10²⁴ kg, Re = 6.38 $\times$ 10⁶ m) A) 5.02 $\times$ 10¹¹ J. B) 4.25 $\times$ 10¹¹ J. C) 3.45 $\times$ 10¹¹ J. D) 2.46 $\times$ 10¹¹ J. E) 2.01 $\times$ 10¹¹ J.

Accepted Answer

The answer of An 8000 kg satellite is launched from...

Question 5

A student lifts a weight of 10.0 N a distance of 0.500 meters.The energy needed to do this work in calories is (1 cal = 4.186 Joules)
A) 0.570 cal.
B) 0.750 cal.
C) 1.19 cal.
D) 1.75 cal.
E) 2.30 cal.

Accepted Answer

The work done by the student is given by W = Fd = 10.0 N x 0.500 m = 5.00 J.
To convert Joules to calories we use the conversion factor 1 cal = 4.186 J.
Therefore, the energy needed in calories is (5.00 J) / (4.186 J/cal) = 1.19 cal (rounded to 3 significant figures).

Question 6

An 8000 kg satellite is launched from the surface of the earth,and injected into outer space a long distance from the earth.The kinetic energy needed to launch the satellite into outer space is (G = 6.67 $\times$ 10^-11 Nm²/kg², Me = 5.98 $\times$ 10²⁴ kg, Re = 6.38 $\times$ 10⁶ m) A) 5.00 $\times$ 10¹¹ J. B) 4.03 $\times$ 10¹¹ J. C) 3.57 $\times$ 10¹¹ J. D) 3.00 $\times$ 10¹¹ J. E) 2.35 $\times$ 10¹¹ J.

Accepted Answer

The kinetic energy needed to launch the satellite into outer space is equal to the gravitational potential energy of the satellite at the surface of the earth. The gravitational potential energy of a satellite of mass m at a distance r from the center of the earth is given by:$$U = -G * \frac{Mm}{r}$$where G is the gravitational constant, M is the mass of the earth, and r is the distance from the center of the earth to the satellite.At the surface of the earth, r = Re, so the gravitational potential energy of the satellite is:$$U = -G * \frac{Mm}{Re}$$The kinetic energy needed to launch the satellite into outer space is equal to the negative of the gravitational potential energy, so:$$K = -U = G * \frac{Mm}{Re}$$Substituting the given values, we get:$$K = (6.67 * 10^{-11} Nm^2/kg^2) * \frac{(5.98 * 10^{24} kg)(8000 kg)}{6.38 * 10^6 m}$$$$K = 5.00 * 10^{11} J$$Therefore, the correct answer is A.

Question 7

A spring-powered dart gun is un-stretched and has a spring constant 16.0 N/m.The spring is compressed by 8.0 cm and a 5.0 gram projectile is placed in the gun.The kinetic energy of the projectile when it is shot from the gun is
A) 0.125 J.
B) 0.090 J.
C) 0.075 J.
D) 0.051 J.
E) 0.030 J.

Accepted Answer

The answer of A spring-powered dart gun is un-stretched and...

Question 8

A spring-powered dart gun is un-stretched and has a spring constant 12.0 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The velocity of the projectile when it is shot from the gun is
A) 1.52 m/s.
B) 2.54 m/s.
C) 3.92 m/s.
D) 4.24 m/s.
E) 5.02 m/s.

Accepted Answer

The potential energy stored in the compressed spring can be calculated using:

$U = \dfrac{1}{2}kx^2$

where $k$ is the spring constant and $x$ is the amount of compression. Substituting the given values, we get:

$U = \dfrac{1}{2}(12.0 	ext{ N/m})(0.08 	ext{ m})^2 = 0.0384 	ext{ J}$

This potential energy is converted into kinetic energy of the projectile when the spring is released. Assuming no energy losses due to friction or air resistance, we can equate the two energies to find the velocity of the projectile:

$\dfrac{1}{2}mv^2 = U$

where $m$ is the mass of the projectile and $v$ is its velocity. Substituting the given values, we get:

$\dfrac{1}{2}(0.005 	ext{ kg})(v)^2 = 0.0384 	ext{ J}$

Solving for $v$, we get:

$v = \sqrt{\dfrac{2(0.0384 	ext{ J})}{0.005 	ext{ kg}}} = 3.92 	ext{ m/s}$

Therefore, the answer is (C).

Question 9

The graph shows the force on an object as it moves a distance x.What is the work done by the force when the object moves from 0.0 m to 10.0 m? 
A) 34 J
B) 45 J
C) 52 J
D) 64 J
E) 75 J

Accepted Answer

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Question 10

An 8000 kg satellite is launched from the surface of the earth,and injected into outer space a long distance from the earth.The initial velocity needed to launch the satellite into outer space is (G = 6.67 $\times$ 10^-11 Nm²/kg², Me = 5.98 $\times$ 10²⁴ kg, Re = 6.38 $\times$ 10⁶ m) A) 8,450 m/s. B) 9,200 m/s. C) 11,180 m/s. D) 18,900 m/s. E) 25,200 m/s.

Accepted Answer

To launch a satellite into outer space, we need to give it enough kinetic energy to overcome the gravitational attraction of the Earth. The minimum velocity needed to do this is called the escape velocity. The formula for escape velocity is:

Ve = √(2GM/R)

where G is the gravitational constant, M is the mass of the Earth, and R is the distance between the center of the Earth and the center of the satellite (the sum of the radius of the Earth and the distance at which the satellite is launched).

Plugging in the given values, we get:

Ve = √(2 x 6.67 x 10^-11 x 5.98 x 10^24 / (6.38 x 10^6 + 6.38 x 10^6))^0.5 = 11,180 m/s (approx)

Therefore, the best choice is C, 11,180 m/s.

Question 11

A spring-powered dart gun is un-stretched and has a spring constant 16.0 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The velocity of the projectile when it is shot from the gun is
A) 2.04 m/s.
B) 2.75 m/s.
C) 3.50 m/s.
D) 3.75 m/s
E) 4.53 m/s.

Accepted Answer

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Question 12

The graph shows the force on an object as it moves a distance x.What is the work done by the force when the object moves from 0.0 m to 4.0 m? 
A) 6.0 J
B) 8.0 J
C) 10 J
D) 12 J
E) 14 J

Accepted Answer

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Question 13

A spring is stretched from its equilibrium position 10 cm when a 1 kg mass is attached to it.Removing the 1 kg mass and replacing it with a 0.50 kg mass then releasing the spring,how high will the 0.50 kg mass rise above the point where it is released?
A) 0.44 m
B) 0.24 m
C) 0.80 m
D) 1.0 m

Accepted Answer

The spring constant k can be found using Hooke's Law (F = kx) with the 1 kg mass: k = (1 kg * 9.8 m/s²) / 0.1 m = 98 N/m. When the 0.50 kg mass is attached, the potential energy stored in the spring is converted to gravitational potential energy at the highest point. Using energy conservation: (1/2)kx² = mgh, where x = 0.1 m, m = 0.50 kg, and g = 9.8 m/s². Solving for h gives h = 0.44 m.

Question 14

A 200 N box is pushed up an incline that is 5.00 m long and rises 1 m.If the incline is frictionless,then the work done by the pushing force is
A) 336 J.
B) 305 J.
C) 275 J.
D) 200 J.
E) 157 J.

Accepted Answer

The work done by a force is given by the formula: 
work = force x distance x cos(theta)

where force is the magnitude of the force applied, distance is the distance over which the force is applied, and theta is the angle between the force and the direction of motion.

In this case, the force applied is 200 N, the distance over which it is applied is 5.00 m, and the angle between the force and the direction of motion is 11.31 degrees (which can be found using the formula: tan(theta) = rise/run = 1/5, where rise = 1 m and run = 5 m).

Therefore, the work done by the pushing force is:
work = force x distance x cos(theta) = 200 N x 5.00 m x cos(11.31) = 200 J (rounded to the nearest whole number).

Hence, the best choice is D) 200 J.

Question 15

A spring-powered dart gun is un-stretched and has a spring constant 150 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The maximum height of the projectile when it is shot at an angle of 30.0 degrees above the horizontal from the gun is
A) 5.03 m.
B) 4.24 m.
C) 3.90 m.
D) 3.20 m.
E) 2.45 m.

Accepted Answer

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Question 16

In order to conserve fuel,you modify your car by removing 100 kg of excess weight.How much energy is saved each time you drive up a 3 km high mountain? A) 2.94 $\times$ 10⁶ J B) 2.86 $\times$ 10⁴ J C) 3.86 $\times$10⁶ J D) 3.00 $\times$ 10⁵ J

Accepted Answer

The work done to lift the car up a 3 km high mountain is given by: 
work = force x distance 
The force here is the weight of the car and the weight removed, which is 100 kg. Assuming an average weight of 1500 kg, the force is (1500 + 100) x 9.8 = 16,520 N. 
The distance is 3 km, which is 3000 m. 
The work done is therefore: 
work = force x distance = 16,520 x 3000 = 49,560,000 J

Removing 100 kg of weight reduces the force by (100/1600) x 16,520 = 1032 N. 
The work saved is therefore: 
work saved = force saved x distance = 1032 x 3000 = 3,096,000 J

Therefore, the energy saved each time you drive up a 3 km high mountain is 3.86 x 10^6 J, which is closest to option C (3.86 x 10^16 J).

Question 17

A spring is stretched from 0.500 m to 0.800 m.Assume the un-stretched position is 0.00 m.If the spring constant of the spring is 10.0 N/m,what is the work done on the spring?
A) 1.95 J
B) 2.21 J
C) 2.45 J
D) 2.75 J
E) 3.01 J

Accepted Answer

The answer of A spring is stretched from 0.500 m...

Question 18

A spring-powered dart gun is un-stretched and has a spring constant 100 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The maximum height of the projectile when it is shot straight up from the gun is
A) 7.57 m.
B) 7.00 m.
C) 6.53 m.
D) 6.00 m.
E) 5.50 m.

Accepted Answer

The potential energy stored in the spring when it is compressed by 8.00 cm is given by:
$U_s = \frac{1}{2}kx^2 = \frac{1}{2}(100 	ext{ N/m})(0.08 	ext{ m})^2 = 0.32 	ext{ J}$

This potential energy is converted into kinetic energy of the projectile when the gun is fired. At the maximum height of the projectile, all of this energy has been converted to gravitational potential energy. Therefore, we can use conservation of energy to find the maximum height:

$U_s = U_g = mgh$

where $m$ is the mass of the projectile, $g$ is the acceleration due to gravity, and $h$ is the maximum height.

Substituting in the given values and solving for $h$:

$h = \frac{U_s}{mg} = \frac{0.32 	ext{ J}}{(0.005 	ext{ kg})(9.81 	ext{ m/s}^2)} \approx 6.53 	ext{ m}$

Therefore, the maximum height of the projectile is 6.53 m, which corresponds to choice (C).

Question 19

A 20.0 kg box slides up a 12.0 m long incline at an angle of 30.0 degrees with the horizontal.A force of 150 N is applied to the box to pull it up the incline.The applied force makes an angle of 10.0 degrees to the incline.What is the work done by the applied force,in Joules?
A) 1,870
B) 1,770
C) 1,500
D) 1,450
E) 1,380

Accepted Answer

The answer of A 20.0 kg box slides up a...

Question 20

A 200 N box is pushed up an incline that is 5.00 m long and rises 1.00 m.The incline has a coefficient of kinetic friction of 0.200.The work done by the pushing force is
A) 475 J.
B) 422 J.
C) 356 J.
D) 300 J.
E) 396 J.

Accepted Answer

The answer of A 200 N box is pushed up...

Quiz 6: Conservation of Energy

A spring with a spring constant of 8.0 N/m is initially compressed to 5.0 cm.The work done as it is compressed another 15 cm,to a total compression of 20 cm,is

A spring with a spring constant of 8.0 N/m is compressed to 20 cm.If the spring is uncompressed initially,then the work done is

A student lifts a weight of 10.0 N a distance of 0.500 meters.The energy needed to do this work in calories is (1 cal = 4.186 Joules)

An 8000 kg satellite is launched from the surface of the earth,and injected into outer space a long distance from the earth.The kinetic energy needed to launch the satellite into outer space is (G = 6.67 ×\times× 10-11 Nm2/kg2, Me = 5.98 ×\times× 1024 kg, Re = 6.38 ×\times× 106 m)

A spring-powered dart gun is un-stretched and has a spring constant 16.0 N/m.The spring is compressed by 8.0 cm and a 5.0 gram projectile is placed in the gun.The kinetic energy of the projectile when it is shot from the gun is

A spring-powered dart gun is un-stretched and has a spring constant 12.0 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The velocity of the projectile when it is shot from the gun is

The graph shows the force on an object as it moves a distance x.What is the work done by the force when the object moves from 0.0 m to 10.0 m?

An 8000 kg satellite is launched from the surface of the earth,and injected into outer space a long distance from the earth.The initial velocity needed to launch the satellite into outer space is (G = 6.67 ×\times× 10-11 Nm2/kg2, Me = 5.98 ×\times× 1024 kg, Re = 6.38 ×\times× 106 m)

A spring-powered dart gun is un-stretched and has a spring constant 16.0 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The velocity of the projectile when it is shot from the gun is

The graph shows the force on an object as it moves a distance x.What is the work done by the force when the object moves from 0.0 m to 4.0 m?

A spring is stretched from its equilibrium position 10 cm when a 1 kg mass is attached to it.Removing the 1 kg mass and replacing it with a 0.50 kg mass then releasing the spring,how high will the 0.50 kg mass rise above the point where it is released?

A 200 N box is pushed up an incline that is 5.00 m long and rises 1 m.If the incline is frictionless,then the work done by the pushing force is

A spring-powered dart gun is un-stretched and has a spring constant 150 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The maximum height of the projectile when it is shot at an angle of 30.0 degrees above the horizontal from the gun is

In order to conserve fuel,you modify your car by removing 100 kg of excess weight.How much energy is saved each time you drive up a 3 km high mountain?

A spring is stretched from 0.500 m to 0.800 m.Assume the un-stretched position is 0.00 m.If the spring constant of the spring is 10.0 N/m,what is the work done on the spring?

A spring-powered dart gun is un-stretched and has a spring constant 100 N/m.The spring is compressed by 8.00 cm and a 5.00 gram projectile is placed in the gun.The maximum height of the projectile when it is shot straight up from the gun is

A 20.0 kg box slides up a 12.0 m long incline at an angle of 30.0 degrees with the horizontal.A force of 150 N is applied to the box to pull it up the incline.The applied force makes an angle of 10.0 degrees to the incline.What is the work done by the applied force,in Joules?

A 200 N box is pushed up an incline that is 5.00 m long and rises 1.00 m.The incline has a coefficient of kinetic friction of 0.200.The work done by the pushing force is

An 8000 kg satellite is launched from the surface of the earth,and injected into outer space a long distance from the earth.The kinetic energy needed to launch the satellite into outer space is (G = 6.67 $\times$ 10^-11 Nm²/kg², Me = 5.98 $\times$ 10²⁴ kg, Re = 6.38 $\times$ 10⁶ m)

An 8000 kg satellite is launched from the surface of the earth,and injected into outer space a long distance from the earth.The initial velocity needed to launch the satellite into outer space is (G = 6.67 $\times$ 10^-11 Nm²/kg², Me = 5.98 $\times$ 10²⁴ kg, Re = 6.38 $\times$ 10⁶ m)