Question 1

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?

Accepted Answer

The energy saved is calculated using the potential energy formula: E = mgh, where m = 100 kg, g = 9.8 m/s², and h = 3000 m. E = 100 * 9.8 * 3000 = 2.94 × 10⁶ J.

Question 2

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 0.0 m to 4.0 m?

Accepted Answer

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

Question 3

An 8000 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? (G = 6.67 × 10⁻¹¹ N·m²/kg², M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m.)

Accepted Answer

The correct answer is A because the initial kinetic energy needed by the satellite is equal to the negative of its gravitational potential energy at infinity, which is given by:$$K = -\frac{GM_Em}{r}$$where:* G is the gravitational constant* M_E is the mass of the Earth* m is the mass of the satellite* r is the distance between the satellite and the center of the EarthAt infinity, r approaches infinity, so the potential energy approaches zero and the kinetic energy becomes:$$K = \frac{GM_Em}{\infty} = 0$$Therefore, the initial kinetic energy needed by the satellite is zero.

Question 4

A 200 N box is pushed at a constant velocity 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, and the pushing force is parallel to the incline. The work done by the pushing force is

Accepted Answer

To find the work done by the pushing force, we need to calculate the force required to move the box up the incline at a constant velocity. This force must overcome both the gravitational component acting down the incline and the frictional force. The gravitational force component (Fg) is given by $mg\sin(\theta)$, where $m$ is the mass, $g$ is the acceleration due to gravity, and $\theta$ is the angle of the incline. The frictional force (Ff) is given by $\mu_k N$, where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force, which is $mg\cos(\theta)$ for an incline. However, without the angle, we can simplify using the given lengths to find the forces because $mg\sin(\theta) = mg(h/l)$ and $N = mg\cos(\theta) = mg\sqrt{1-(h/l)^2}$ for small angles or directly as $mg$ for the vertical component when ignoring the angle simplification due to the lack of direct angle information.Given:- $m = 200\,N/g$ (since weight is given, and weight = $mg$)- $g = 9.8\,m/s^2$- $\mu_k = 0.200$- Incline length = 5.00 m- Height = 1.00 mFirst, calculate the gravitational force component acting down the incline:$$Fg = mg\frac{h}{l} = 200\,N \times \frac{1}{5} = 40\,N$$Then, calculate the frictional force:$$Ff = \mu_k \times mg = 0.200 \times 200\,N = 40\,N$$The total force required to move the box up at a constant velocity is the sum of these two forces:$$F_{total} = Fg + Ff = 40\,N + 40\,N = 80\,N$$Finally, the work done by this force over the distance of the incline is:$$W = F_{total} \times d = 80\,N \times 5\,m = 400\,J$$However, this calculation does not directly account for the work against gravity, which is already included in the calculation of the gravitational force component acting down the incline. The correct calculation involves recognizing that the total work done includes overcoming both gravity and friction over the distance of the incline. The slight discrepancy in the final answer from the provided options suggests a reevaluation of the calculation process or an approximation in the options. Given the closest option to the calculated work and considering potential rounding or simplification in the problem setup, option E (396 J) is selected as the closest match.

Question 5

A spring-powered dart gun is unstretched 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

Accepted Answer

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

Question 6

An 8000 kg satellite is launched from the surface of the Earth into outer space. What initial speed 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? (G = 6.67 × 10⁻¹¹ N·m²/kg², M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m.)

Accepted Answer

We can use the conservation of energy principle to solve the problem:
Initial energy = Final energy
Initial kinetic energy + Initial potential energy = Final potential energy
1/2 mv^2 - GmM/R = -GmM/∞
where m = 8000 kg, M = 5.97 × 10^24 kg, R = 6.37 × 10^6 m, and ∞ is infinity.
Solving for v, we get:
v = √(2GM/R) = √(2(6.67 × 10^-11)(5.97 × 10^24)/(6.37 × 10^6)) = 11,180 m/s
Therefore, the answer is choice C.

Question 7

A spring-powered dart gun is unstretched 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

Accepted Answer

The potential energy stored in the spring when it is compressed is given by:
$U_s = \frac{1}{2}kx^2$
where k is the spring constant and x is the compression of the spring.
Substituting the given values, we have:
$U_s = \frac{1}{2}(100 N/m)(0.08 m)^2 = 0.32 J$
This potential energy is converted to kinetic energy of the projectile when it is released. The kinetic energy of the projectile is given by:
$K = \frac{1}{2}mv^2$
where m is the mass of the projectile and v is its velocity.
We can use conservation of energy to find the velocity of the projectile at the instant it leaves the gun. At the maximum height, the kinetic energy of the projectile is zero and all the initial potential energy is converted to gravitational potential energy:
$U_g = mgh$
where h is the maximum height reached by the projectile.
We can equate the initial potential energy to the final gravitational potential energy:
$U_s = U_g$
$\frac{1}{2}kx^2 = mgh$
Solving for h, we get:
$h = \frac{1}{2}(\frac{kx^2}{mg})$
Substituting the given values, we have:
$h = \frac{1}{2}(\frac{(100 N/m)(0.08 m)^2}{(0.005 kg)(9.81 m/s^2)}) = 6.53 m$
Therefore, the maximum height of the projectile when it is shot straight up from the gun is 6.53 m. The answer is choice (C).

Question 8

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

Accepted Answer

The work done on the spring is calculated using the formula for the work done in compressing or stretching a spring: $W = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. Substituting $k = 8.0 \, \text{N/m}$ and $x = 0.20 \, \text{m}$ (20 cm = 0.20 m), we get $W = \frac{1}{2} \times 8.0 \, \text{N/m} \times (0.20 \, \text{m})^2 = 0.16 \, \text{J}$.

Question 9

A 1 kg mass is attached to a vertical spring, stretching it by 10 cm from its equilibrium position. With the spring secured in the same stretched position, the 1 kg mass is removed and replaced with a 0.50 kg mass. If the spring is then released, how high will the 0.50 kg mass rise above the point where it was released?

Accepted Answer

The potential energy stored in the spring is equal to the work done in stretching it. When the 1 kg mass is attached, the work done is:$$W = Fd = mgd = (1 kg)(9.8 m/s^2)(0.10 m) = 0.98 J$$When the 0.50 kg mass is attached, the work done is:$$W = Fd = mgd = (0.50 kg)(9.8 m/s^2)(d) = 4.9d J$$Since the spring is secured in the same stretched position, the work done in both cases is the same. Therefore, we have:$$0.98 J = 4.9d J$$Solving for d, we get:$$d = 0.20 m$$Therefore, the 0.50 kg mass will rise to a height of 0.20 m above the point where it was released.

Question 10

A spring is stretched from 0.500 m to 0.800 m. Assume the unstretched 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?

Accepted Answer

Work done on the spring is given by the formula: W = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, k = 10.0 N/m and x = 0.800 m - 0.500 m = 0.300 m. Therefore, W = (1/2)(10.0 N/m)(0.300 m)^2 = 1.95 J.

Question 11

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 × 10⁻¹¹ N·m²/kg², M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m)

Accepted Answer

The gravitational potential energy of an object in orbit can be found using the formula:

U = -G*M*m/r

where U is the gravitational potential energy, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the centers of the two objects.

In this problem, we are given that the mass of the satellite is 8000 kg, the altitude of the orbit is 100 km (which gives us the distance between the centers of the Earth and the satellite), and the values of G, M, and R are also given.

Using the formula and plugging in the values, we get:

U = -6.67 × 10^-11 * 5.97 × 10^24 * 8000 / (6.37 × 10^6 + 100000)

U = -4.92 × 10^11 J

Therefore, the gravitational potential energy of the satellite in the circular orbit is -4.92 × 10^11 J. The best choice is B.

Question 12

A 200 N box is pushed at a constant velocity 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

Accepted Answer

Since the box is moving at a constant velocity, the net force on the box must be zero. This means that the force of gravity (weight) pulling the box down the incline is exactly offset by the force pushing the box up the incline. The work done by the pushing force is given by the distance it moves the box, multiplied by the component of the pushing force that is parallel to the incline. In this case, the pushing force is exactly equal and opposite to the weight of the box, so the magnitude of the parallel component of the pushing force is just the weight of the box times the sine of the angle of the incline (which is 1/5). Thus, the work done by the pushing force is:

W = F_parallel * d = (200 N) * sin(1/5) * 5.00 m = 200 J

Therefore, the answer is D) 200 J.

Question 13

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)

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 14

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 0.0 m to 10.0 m?

Accepted Answer

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

Question 15

A spring-powered dart gun is unstretched 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

Accepted Answer

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

Question 16

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 × 10⁻¹¹ N·m²/kg², M_E = 5.97 × 1024 kg, R_E = 6.37 × 10⁶ m)

Accepted Answer

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

Question 17

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

Accepted Answer

The answer of A spring with a spring constant of...

Question 18

A spring-powered dart gun is unstretched 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

Accepted Answer

The velocity of the projectile can be found using the conservation of energy principle. The potential energy stored in the compressed spring is converted into the kinetic energy of the projectile. The potential energy ($PE$) stored in the spring when compressed is given by the formula $PE = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the compression distance. The kinetic energy ($KE$) of the projectile when it is shot is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the projectile and $v$ is its velocity. Setting the potential energy equal to the kinetic energy and solving for $v$ gives:$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$Given $k = 12.0\, N/m$, $x = 8.00\, cm = 0.08\, m$, and $m = 5.00\, g = 0.005\, kg$, we can solve for $v$:$$12.0 \times 0.08^2 = 0.005 \times v^2$$$$v^2 = \frac{12.0 \times 0.08^2}{0.005}$$$$v = \sqrt{\frac{12.0 \times 0.08^2}{0.005}} \approx 3.92\, m/s$$Therefore, the velocity of the projectile is approximately 3.92 m/s.

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?

Accepted Answer

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

Question 20

A spring-powered dart gun is unstretched 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

Accepted Answer

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

Quiz 6: Conservation of Energy

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?

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 0.0 m to 4.0 m?

A 200 N box is pushed at a constant velocity 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, and the pushing force is parallel to the incline. The work done by the pushing force is

A spring-powered dart gun is unstretched 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 unstretched 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 spring with a spring constant of 8.0 N/m is compressed to 20 cm. If the spring was uncompressed initially, then the work done (by the compressing force) is

A spring is stretched from 0.500 m to 0.800 m. Assume the unstretched 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 200 N box is pushed at a constant velocity 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 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)

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 0.0 m to 10.0 m?

A spring-powered dart gun is unstretched 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

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 × 10−11 N·m2/kg2, ME = 5.97 × 1024 kg, RE = 6.37 × 106 m)

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

A spring-powered dart gun is unstretched 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 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?

A spring-powered dart gun is unstretched 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

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 × 10⁻¹¹ N·m²/kg², M_E = 5.97 × 1024 kg, R_E = 6.37 × 10⁶ m)