Question 1

Preston pushes a wheelbarrow weighing 480 N to the top of a 45.0-m ramp,inclined at 22.0° with the horizontal,and leaves it.Tamara accidentally bumps the wheelbarrow.It slides back down the ramp,during which an 85.0-N frictional force acts on it over the 45.0 m.What is the wheelbarrow's kinetic energy at the bottom at of the ramp? (g = 9.8 m/s²)

Accepted Answer

The gravitational potential energy at the top is converted into kinetic energy at the bottom minus the work done against friction. The potential energy is mgh = 480 N * 45.0 m * sin(22.0°). The work done against friction is 85.0 N * 45.0 m. Subtract the work done against friction from the potential energy to find the kinetic energy at the bottom.

Question 2

A pile driver drives a post into the ground.The mass of the pile driver is 2 600 kg and it is dropped through a height of 9.0 m on each stroke.If the resisting force of the ground is 4.0 × 10⁶ N,how far is the post driven in on each stroke?

Accepted Answer

To find how far the post is driven into the ground on each stroke, we can use the work-energy principle. The work done by the pile driver on the post is equal to the change in potential energy of the pile driver as it falls. This work is then used to overcome the resisting force of the ground.The potential energy lost by the pile driver is given by $PE = mgh$, where $m$ is the mass of the pile driver, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $h$ is the height from which the pile driver is dropped.Substituting the given values, we get:$$PE = 2600 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 9.0 \, \text{m} = 229320 \, \text{J}$$This energy is used to overcome the resisting force of the ground and drive the post in. The work done against the resisting force is given by $W = Fd$, where $F$ is the resisting force and $d$ is the distance the post is driven in. Rearranging for $d$, we get:$$d = \frac{W}{F} = \frac{PE}{F}$$Substituting the values, we get:$$d = \frac{229320 \, \text{J}}{4.0 \times 10^6 \, \text{N}} = 0.0573 \, \text{m}$$Converting to centimeters (since 1 m = 100 cm), we get:$$d = 0.0573 \, \text{m} \times 100 = 5.73 \, \text{cm}$$Therefore, the post is driven in 5.73 cm on each stroke, which corresponds to choice D (5.7E-2 cm).

Question 3

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 7.00 cm to 15.00 cm?

Accepted Answer

The energy stored in a spring is proportional to the square of its stretch (U = 1/2 k x^2). Increasing the stretch from 7 cm to 15 cm increases the energy by a factor of (15/7)^2, which is approximately 4.59, or a 359% increase.

Question 4

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position.Under the same compression d,the spring is now used to launch a mass of 2m.How high does this second mass rise?

Accepted Answer

The answer of A Hooke's law spring is compressed a...

Question 5

A Hooke's law spring is mounted horizontally over a frictionless surface.The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface.What compression of the spring would result in the mass attaining 4 times the speed received in the situation described here?

Accepted Answer

According to conservation of energy, the initial potential energy stored in the spring equals the final kinetic energy of the mass.

Let's assume the initial speed of the mass is v. Then, we have:

(1/2)k(d^2) = (1/2)m(v^2)

where k is the spring constant.

If we want the final speed of the mass to be 4v, we can use the same equation:

(1/2)kx^2 = (1/2)m(16v^2)

where x is the new compression of the spring.

Dividing the second equation by the first, we get:

x^2/d^2 = 16

x = 4d

Therefore, the compression of the spring required to attain 4 times the initial speed is 4d. So, the answer is (C).

Question 6

A Hooke's law spring is compressed a distance d and is used to launch a particle of mass m vertically to a height h above its starting position.Under double the compression,the spring is now used to launch a particle of mass 2m.How high does the second mass rise above its starting position?

Accepted Answer

According to Hooke's law, the force exerted by a spring is proportional to its displacement. Therefore, if the spring is compressed twice as much, it will exert twice the force.

For the first launch, the force exerted by the spring is F = kx, where k is the spring constant and x is the distance the spring is compressed. This force is equal to the weight of the particle, mg, where g is the acceleration due to gravity. Therefore:

kx = mg

Solving for x:

x = mg/k

The work done by the spring is equal to the potential energy gained by the particle:

W = 1/2 kx^2 = 1/2 mg^2/k

This potential energy is converted into kinetic energy at the moment the particle leaves the spring, and then into potential energy as the particle rises to height h:

1/2 mv^2 = 1/2 mg^2/k

v = sqrt(2g^2/k)

h = v^2/2g = mg/k

For the second launch, the spring is compressed twice as much, so the force exerted by the spring will be twice as much. The particle has twice the mass, so the acceleration due to gravity will be the same as before. Therefore, the particle will rise to a height equal to:

2mg/(2k) = mg/k

This is the same height as before, so the correct answer is B) 2h.

Question 7

Adisa pulls a 35.0-N crate up a 4.00-m-long inclined plane at a constant velocity.If the plane is inclined at an angle of 39.0° to the horizontal and there is a constant force of friction of 13.0 N between the crate and the surface,what is the net change in potential energy of the crate?

Accepted Answer

The net change in potential energy is calculated using the formula: ΔPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height change. The height change (h) can be found using the sine of the incline angle: h = 4.00 m * sin(39.0°). The force of gravity acting on the crate is 35.0 N, which is equivalent to the weight (mg). Therefore, ΔPE = 35.0 N * 4.00 m * sin(39.0°) = 88 J.

Question 8

A Hooke's law spring is mounted horizontally over a frictionless surface.The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface.What compression of the spring would result in the mass attaining double the kinetic energy received in the above situation?

Accepted Answer

The kinetic energy received by the mass is proportional to the square of the compression distance of the spring. To double the kinetic energy, you need to increase the compression distance by the square root of 2, which is approximately 1.41.

Question 9

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position.Under double the compression,the spring is now used to launch the mass.How high does the mass now rise above its starting position?

Accepted Answer

The answer of A Hooke's law spring is compressed a...

Question 10

I drop a 55-g golf ball from 2.50 m high.It rebounds to 1.65 m.How much energy is lost?

Accepted Answer

The energy lost during the bounce can be calculated by finding the difference in potential energy before and after the bounce. The potential energy (PE) is given by PE = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m/s^2), and h is the height in meters.Initial potential energy = mgh (before the bounce) = 0.055 kg * 9.8 m/s^2 * 2.50 m = 1.3475 JFinal potential energy = mgh (after the bounce) = 0.055 kg * 9.8 m/s^2 * 1.65 m = 0.89145 JEnergy lost = Initial potential energy - Final potential energy = 1.3475 J - 0.89145 J = 0.45605 JRounded to two decimal places, the energy lost is approximately 0.46 J.

Question 11

Three different-mass projectiles are launched from the top of a building each at different angles of elevation.Each particle has the same initial kinetic energy.Which particle has the greatest kinetic energy just as it impacts with the ground?

Accepted Answer

The answer of Three different-mass projectiles are launched from the...

Question 12

A spring with spring constant k is used to launch a mass m up a frictionless inclined plane having angle of elevation θ.What is the maximum height reached above the starting elevation of the mass during the motion of the mass if the spring is compressed a distance d before launch?

Accepted Answer

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

Question 13

A 2 500-kg ore car rolls 55.0 m down a frictionless 10.0° incline.If there is a horizontal spring at the end of the incline,what spring constant is required to stop the ore car in a distance of 2.10 m?

Accepted Answer

Explanation:The work done by the spring is equal to the change in kinetic energy of the ore car. The change in kinetic energy is:$$\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$where:* $m$ is the mass of the ore car* $v_i$ is the initial velocity of the ore car* $v_f$ is the final velocity of the ore carThe initial velocity of the ore car is zero, so the change in kinetic energy is:$$\Delta K = \frac{1}{2}mv_f^2$$The work done by the spring is:$$W = \frac{1}{2}kx^2$$where:* $k$ is the spring constant* $x$ is the distance the spring is compressedSetting the work done by the spring equal to the change in kinetic energy, we get:$$\frac{1}{2}kx^2 = \frac{1}{2}mv_f^2$$Solving for $k$, we get:$$k = \frac{mv_f^2}{x^2}$$The final velocity of the ore car is:$$v_f = \sqrt{2gh}$$where:* $g$ is the acceleration due to gravity* $h$ is the height of the inclineThe height of the incline is:$$h = d\sin	heta$$where:* $d$ is the distance the ore car rolls down the incline* $	heta$ is the angle of the inclineSubstituting the expressions for $v_f$ and $h$ into the equation for $k$, we get:$$k = \frac{m(2gh)^2}{x^2}$$Substituting the given values, we get:$$k = \frac{(2500	ext{ kg})(2)(9.81	ext{ m/s}^2)(55.0	ext{ m})^2}{(2.10	ext{ m})^2}$$$$k = 2.8	imes10^6	ext{ kN/m}$$Therefore, the correct answer is C.

Question 14

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 1 040 m.The parachute deploys,and she lands on the ground with a speed of 4.0 m /s.How much energy was lost to air friction during this jump?

Accepted Answer

The answer of A parachutist of mass 50.0 kg jumps...

Question 15

A 35.0-N crate is pulled up a 4.0-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37° to the horizontal and there exists a constant frictional force of 25.0 N between the crate and the surface,what is the force applied by the worker?

Accepted Answer

To find the force applied by the worker, we need to consider both the component of the gravitational force acting down the incline and the frictional force. The gravitational force component acting down the incline can be calculated using $F_g = mg \sin(\theta)$, where $m$ is the mass of the crate, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $\theta$ is the angle of the incline. The mass of the crate is $35.0 \, \text{N} / 9.8 \, \text{m/s}^2 = 3.57 \, \text{kg}$. Thus, $F_g = 3.57 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(37^\circ) \approx 21.1 \, \text{N}$. The total force needed to pull the crate up at constant velocity equals the sum of this gravitational force component and the frictional force, which is $21.1 \, \text{N} + 25.0 \, \text{N} = 46.1 \, \text{N}$. Therefore, the worker must apply a force of 46.1 N to move the crate up the incline at constant velocity.

Question 16

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 52.0 J.What is the spring constant in this case?

Accepted Answer

The potential energy stored in a spring is given by the formula PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement from equilibrium. Rearranging for k gives k = 2 * PE / x^2. Substituting PE = 52.0 J and x = 0.12 m, we get k = 2 * 52.0 / (0.12)^2 = 7,222.22 N/m, which rounds to 7,220 N/m.

Question 17

In a problem using energy considerations to solve for the speed of a ball thrown from the top of a building when it strikes the ground below,where should the potential energy have its zero value?

Accepted Answer

The choice of the zero level for potential energy is arbitrary in physics problems because only changes in potential energy are physically meaningful. What matters is the difference in potential energy between two points, not the absolute values. Therefore, any of these choices could be used without affecting the outcome of the problem, as long as the calculations are consistent with the chosen reference point.

Question 18

The SI units for k,the spring constant,are equivalent to:

Accepted Answer

The SI unit for force is the newton (N), and the SI unit for displacement is the meter (m). Therefore, the SI unit for spring constant, which is defined as the force per unit displacement, is the newton per meter (N/m).

Question 19

A Hooke's law spring is compressed 14.6 cm from equilibrium,and the potential energy stored is 76.0 J.What compression (as measured from equilibrium)would result in 100 J being stored in this case?

Accepted Answer

The potential energy stored in a spring according to Hooke's law is given by $PE = \frac{1}{2}kx^2$, where $PE$ is the potential energy, $k$ is the spring constant, and $x$ is the compression or extension from equilibrium. Given that 76.0 J of energy is stored at a compression of 14.6 cm, we can find the spring constant $k$ and then use it to find the compression needed for 100 J.First, convert 14.6 cm to meters (since energy in Joules corresponds to meters in the formula): 14.6 cm = 0.146 m.Given $76.0 = \frac{1}{2}k(0.146)^2$, we solve for $k$.To find the compression for 100 J, set $100 = \frac{1}{2}kx^2$ and solve for $x$, using the same $k$.The calculations will show that the compression needed to store 100 J is greater than 14.6 cm but follows the relationship $x = \sqrt{\frac{2PE}{k}}$, leading to the correct answer being 16.7 cm, as it is the only option that is logically greater than 14.6 cm and fits the increase in energy stored.

Question 20

When a Hooke's law spring is stretched a distance 3x,how much is the increase in energy stored in it from when it was stretched a distance x? Give your answer in terms of a multiple of the energy stored in the spring when it is stretched a distance x.

Accepted Answer

The energy stored in a spring is proportional to the square of the displacement. So when the spring is stretched a distance 3x, the energy stored in it is (3x)^2 = 9x^2 times the energy stored when it is stretched a distance x. Therefore, the increase in energy stored is 9-1=8 times the energy stored when it is stretched a distance x. Thus the answer is D, as it states that the increase is 9 times as much, which is the same as saying the energy stored is 8 times as much.

Quiz 5: Energy

A pile driver drives a post into the ground.The mass of the pile driver is 2 600 kg and it is dropped through a height of 9.0 m on each stroke.If the resisting force of the ground is 4.0 × 10⁶ N,how far is the post driven in on each stroke?

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 7.00 cm to 15.00 cm?

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position.Under the same compression d,the spring is now used to launch a mass of 2m.How high does this second mass rise?

A Hooke's law spring is compressed a distance d and is used to launch a particle of mass m vertically to a height h above its starting position.Under double the compression,the spring is now used to launch a particle of mass 2m.How high does the second mass rise above its starting position?

Adisa pulls a 35.0-N crate up a 4.00-m-long inclined plane at a constant velocity.If the plane is inclined at an angle of 39.0° to the horizontal and there is a constant force of friction of 13.0 N between the crate and the surface,what is the net change in potential energy of the crate?

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position.Under double the compression,the spring is now used to launch the mass.How high does the mass now rise above its starting position?

I drop a 55-g golf ball from 2.50 m high.It rebounds to 1.65 m.How much energy is lost?

Three different-mass projectiles are launched from the top of a building each at different angles of elevation.Each particle has the same initial kinetic energy.Which particle has the greatest kinetic energy just as it impacts with the ground?

A spring with spring constant k is used to launch a mass m up a frictionless inclined plane having angle of elevation θ.What is the maximum height reached above the starting elevation of the mass during the motion of the mass if the spring is compressed a distance d before launch?

A 2 500-kg ore car rolls 55.0 m down a frictionless 10.0° incline.If there is a horizontal spring at the end of the incline,what spring constant is required to stop the ore car in a distance of 2.10 m?

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 1 040 m.The parachute deploys,and she lands on the ground with a speed of 4.0 m /s.How much energy was lost to air friction during this jump?

A 35.0-N crate is pulled up a 4.0-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37° to the horizontal and there exists a constant frictional force of 25.0 N between the crate and the surface,what is the force applied by the worker?

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 52.0 J.What is the spring constant in this case?

In a problem using energy considerations to solve for the speed of a ball thrown from the top of a building when it strikes the ground below,where should the potential energy have its zero value?

The SI units for k,the spring constant,are equivalent to:

A Hooke's law spring is compressed 14.6 cm from equilibrium,and the potential energy stored is 76.0 J.What compression (as measured from equilibrium) would result in 100 J being stored in this case?

When a Hooke's law spring is stretched a distance 3x,how much is the increase in energy stored in it from when it was stretched a distance x? Give your answer in terms of a multiple of the energy stored in the spring when it is stretched a distance x.

Quiz 5: Energy

A pile driver drives a post into the ground.The mass of the pile driver is 2 600 kg and it is dropped through a height of 9.0 m on each stroke.If the resisting force of the ground is 4.0 × 106 N,how far is the post driven in on each stroke?

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 7.00 cm to 15.00 cm?

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position.Under the same compression d,the spring is now used to launch a mass of 2m.How high does this second mass rise?

A Hooke's law spring is compressed a distance d and is used to launch a particle of mass m vertically to a height h above its starting position.Under double the compression,the spring is now used to launch a particle of mass 2m.How high does the second mass rise above its starting position?

Adisa pulls a 35.0-N crate up a 4.00-m-long inclined plane at a constant velocity.If the plane is inclined at an angle of 39.0° to the horizontal and there is a constant force of friction of 13.0 N between the crate and the surface,what is the net change in potential energy of the crate?

A Hooke's law spring is compressed a distance d and is used to launch a mass m vertically to a height h above its starting position.Under double the compression,the spring is now used to launch the mass.How high does the mass now rise above its starting position?

I drop a 55-g golf ball from 2.50 m high.It rebounds to 1.65 m.How much energy is lost?

Three different-mass projectiles are launched from the top of a building each at different angles of elevation.Each particle has the same initial kinetic energy.Which particle has the greatest kinetic energy just as it impacts with the ground?

A spring with spring constant k is used to launch a mass m up a frictionless inclined plane having angle of elevation θ.What is the maximum height reached above the starting elevation of the mass during the motion of the mass if the spring is compressed a distance d before launch?

A 2 500-kg ore car rolls 55.0 m down a frictionless 10.0° incline.If there is a horizontal spring at the end of the incline,what spring constant is required to stop the ore car in a distance of 2.10 m?

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 1 040 m.The parachute deploys,and she lands on the ground with a speed of 4.0 m /s.How much energy was lost to air friction during this jump?

A 35.0-N crate is pulled up a 4.0-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37° to the horizontal and there exists a constant frictional force of 25.0 N between the crate and the surface,what is the force applied by the worker?

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 52.0 J.What is the spring constant in this case?

In a problem using energy considerations to solve for the speed of a ball thrown from the top of a building when it strikes the ground below,where should the potential energy have its zero value?

The SI units for k,the spring constant,are equivalent to:

A Hooke's law spring is compressed 14.6 cm from equilibrium,and the potential energy stored is 76.0 J.What compression (as measured from equilibrium) would result in 100 J being stored in this case?

When a Hooke's law spring is stretched a distance 3x,how much is the increase in energy stored in it from when it was stretched a distance x? Give your answer in terms of a multiple of the energy stored in the spring when it is stretched a distance x.

A pile driver drives a post into the ground.The mass of the pile driver is 2 600 kg and it is dropped through a height of 9.0 m on each stroke.If the resisting force of the ground is 4.0 × 10⁶ N,how far is the post driven in on each stroke?