Question 1

I drop a 60-g golf ball from 2.0 m high.It rebounds to 1.5 m.How much energy is lost?

Accepted Answer

The potential energy at 2.0 m is mgh = 0.06 kg * 9.8 m/s² * 2.0 m = 1.176 J. The potential energy at 1.5 m is 0.06 kg * 9.8 m/s² * 1.5 m = 0.882 J. The energy lost is 1.176 J - 0.882 J = 0.294 J, which rounds to 0.29 J.

Question 2

A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37 $\degree$ to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface,what is the force applied by the worker?

Accepted Answer

The answer of A 50-N crate is pulled up a...

Question 3

Adisa pulls a 40-N crate up a 5.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 37 $\degree$ to the horizontal and there is a constant force of friction of 10 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: PE = mgh. The height (h) can be found using the sine of the incline angle: h = 5.0 m * sin(37°). The weight (mg) is 40 N. Thus, PE = 40 N * 5.0 m * sin(37°) = 120 J.

Question 4

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 8.00 cm to 16.0 cm?

Accepted Answer

The energy stored in a Hooke's law spring is given by the formula U = (1/2)kx^2, where k is the spring constant and x is the displacement from equilibrium. Since the displacement is proportional to the stretch, we can ignore the units and use the ratio of the stretches to determine the change in energy.

The initial stretch is 8.00 cm and the final stretch is 16.0 cm, which is a factor of 2. Therefore, the energy stored in the spring is increased by a factor of (2^2) = 4. This is equivalent to a 300% increase, so the answer is (C).

Question 5

A girl and her bicycle have a total mass of 40 kg.At the top of the hill her speed is 5.0 m/s.The hill is 10 m high and 100 m long.If the force of friction as she rides down the hill is 20 N,what is her speed at the bottom?

Accepted Answer

The initial potential energy of the girl and her bicycle at the top of the hill is given by:
PE1 = mgh = 40 kg × 9.81 m/s² × 10 m = 3,924 J

This potential energy is converted into kinetic energy at the bottom of the hill, where the speed of the girl and her bicycle is maximum. The final kinetic energy is given by:
KE2 = (1/2)mv²

The loss of energy due to friction along the way is given by:
ΔE = Wfriction = Ffriction × d = 20 N × 100 m = 2,000 J

Therefore, the final kinetic energy can be found by subtracting the loss of energy due to friction from the initial potential energy:
KE2 = PE1 - ΔE = 3,924 J - 2,000 J = 1,924 J

Solving for the final velocity v yields:
KE2 = (1/2)mv²
1,924 J = (1/2)(40 kg)v²
v = sqrt(2 × 1,924 J / 40 kg) = 10.9 m/s ≈ 11 m/s

Therefore, the answer is C)11 m/s.

Question 6

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 speed received in the above situation?

Accepted Answer

The answer of A Hooke's law spring is mounted horizontally...

Question 7

A bobsled makes a run down an ice track starting at 150 m vertical distance up the hill.If there is no friction,what is the velocity at the bottom of the hill?

Accepted Answer

The answer of A bobsled makes a run down an...

Question 8

A pile driver drives a post into the ground.The mass of the pile driver is 2 500 kg and it is dropped through a height of 8.0 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

The potential energy lost by the pile driver is converted into work done against the resisting force. The potential energy is mgh = 2500 kg * 9.8 m/s² * 8.0 m = 196,000 J. The work done is force * distance, so 4.0 * 10^6 N * distance = 196,000 J. Solving for distance gives 0.049 m or 4.9 cm.

Question 9

An amount of work equal to 1.5 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 15-g marble?

Accepted Answer

The potential energy stored in the spring is converted into the kinetic energy of the marble. Using the formula for kinetic energy (KE = 0.5 * m * v^2) and setting it equal to the potential energy (1.5 J), we solve for v: 1.5 = 0.5 * 0.015 * v^2, which gives v ≈ 14 m/s.

Question 10

A girl and her bicycle have a total mass of 40.0 kg.At the top of the hill her speed is 5.0 m/s,and her speed doubles as she rides down the hill.The hill is 10.0 m high and 100 m long.How much kinetic energy and potential energy is lost to friction?

Accepted Answer

Initial potential energy = mgh = 40.0 kg x 9.8 m/s^2 x 10.0 m = 3 920 J
Initial kinetic energy = 0.5 x 40.0 kg x (5.0 m/s)^2 = 500 J

Final potential energy = mgh = 40.0 kg x 9.8 m/s^2 x 0 m = 0 J
Final kinetic energy = 0.5 x 40.0 kg x (2 x 5.0 m/s)^2 = 2 000 J

The total mechanical energy lost to friction is the difference between the initial mechanical energy and the final mechanical energy:
ΔE = (initial potential energy + initial kinetic energy) - (final potential energy + final kinetic energy)
ΔE = (3 920 J + 500 J) - (0 J + 2 000 J)
ΔE = 2 420 J

Therefore, the answer is (A) 2 420 J.

Question 11

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 12

A Hooke's law spring is compressed 12.0 cm from equilibrium,and the potential energy stored is 72.0 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 is given by $ PE = \frac{1}{2} k x^2 $. Using the given data, we can find the spring constant $ k $ and then solve for the new compression $ x $ that results in 100 J of potential energy.

Question 13

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J.What is the spring constant in this case?

Accepted Answer

The potential energy stored in a spring is given by the equation:$$U = \frac{1}{2}kx^2$$where:* U is the potential energy stored in the spring (in joules)* k is the spring constant (in newtons per meter)* x is the displacement of the spring from equilibrium (in meters)In this case, we are given that the potential energy stored in the spring is 72.0 J and the displacement of the spring from equilibrium is 12.0 cm (or 0.12 m). Substituting these values into the equation above, we get:$$72.0 J = \frac{1}{2}k(0.12 m)^2$$Solving for k, we get:$$k = \frac{2(72.0 J)}{(0.12 m)^2} = 10,000 N/m$$Therefore, the spring constant in this case is 10,000 N/m.

Question 14

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

Accepted Answer

The answer of The SI units for k,the spring constant,are...

Question 15

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 1 000 m.The parachute deploys,and she lands on the ground with a speed of 5.0 m/s.How much energy was lost to air friction during this jump?

Accepted Answer

The potential energy at the start is mgh = 50 * 9.8 * 1000 = 490,000 J. The kinetic energy when landing is 0.5 * 50 * 5^2 = 625 J. The energy lost to air friction is the difference: 490,000 J - 625 J = 489,375 J, which rounds to 489,000 J.

Question 16

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 17

A baseball catcher puts on an exhibition by catching a 0.150-kg ball dropped from a helicopter at a height of 100 m above the catcher.If the catcher "gives" with the ball for a distance of 0.750 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

Accepted Answer

The potential energy of the ball at 100 m is converted to kinetic energy just before the catch. Using energy conservation, the velocity just before the catch is found. Then, using work-energy principle, the average force exerted is calculated considering the distance over which the force acts.

Question 18

A 2 000-kg ore car rolls 50.0 m down a frictionless 10.0 $\degree$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 1.00 m?

Accepted Answer

The work done by the spring on the ore car 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 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:$$\frac{1}{2}kx^2 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$Solving for $k$, we get:$$k = \frac{mv_f^2 - mv_i^2}{x^2}$$Plugging in the given values, we get:$$k = \frac{(2000 	ext{ kg})(0 	ext{ m/s})^2 - (2000 	ext{ kg})(10 	ext{ m/s})^2}{(1.00 	ext{ m})^2}$$$$k = 340 	ext{ kN/m}$$

Question 19

A 20-N crate starting at rest slides down a rough 5.0-m long ramp,inclined at 25 $\degree$ with the horizontal.20 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?

Accepted Answer

The answer of A 20-N crate starting at rest slides...

Question 20

Preston pushes a wheelbarrow weighing 500 N to the top of a 50.0-m ramp,inclined at 20.0 $\degree$ with the horizontal,and leaves it.Tamara accidentally bumps the wheelbarrow.It slides back down the ramp,during which an 80.0-N frictional force acts on it over the 50.0 m.What is the wheelbarrow's kinetic energy at the bottom at of the ramp? (g = 9.8 m/s²)

Accepted Answer

Explanation:The work done by Preston in pushing the wheelbarrow up the ramp is:$$W = Fd\cos	heta = (500	ext{ N})(50.0	ext{ m})\cos20.0^\circ = 45.5	ext{ kJ}$$The work done by friction in sliding back down the ramp is:$$W = Fd = (80.0	ext{ N})(50.0	ext{ m}) = 4.00	ext{ kJ}$$The net work done on the wheelbarrow is:$$W = 45.5	ext{ kJ} - 4.00	ext{ kJ} = 41.5	ext{ kJ}$$The kinetic energy of the wheelbarrow at the bottom of the ramp is:$$K = W = 41.5	ext{ kJ} = \boxed{4150	ext{ J}}$$

Quiz 5: Energy

I drop a 60-g golf ball from 2.0 m high.It rebounds to 1.5 m.How much energy is lost?

A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37 $\degree$ to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface,what is the force applied by the worker?

Adisa pulls a 40-N crate up a 5.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 37 $\degree$ to the horizontal and there is a constant force of friction of 10 N between the crate and the surface,what is the net change in potential energy of the crate?

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 8.00 cm to 16.0 cm?

A girl and her bicycle have a total mass of 40 kg.At the top of the hill her speed is 5.0 m/s.The hill is 10 m high and 100 m long.If the force of friction as she rides down the hill is 20 N,what is her speed at the bottom?

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 speed received in the above situation?

A bobsled makes a run down an ice track starting at 150 m vertical distance up the hill.If there is no friction,what is the velocity at the bottom of the hill?

A pile driver drives a post into the ground.The mass of the pile driver is 2 500 kg and it is dropped through a height of 8.0 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?

An amount of work equal to 1.5 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 15-g marble?

A girl and her bicycle have a total mass of 40.0 kg.At the top of the hill her speed is 5.0 m/s,and her speed doubles as she rides down the hill.The hill is 10.0 m high and 100 m long.How much kinetic energy and potential energy is lost to friction?

A Hooke's law spring is compressed 12.0 cm from equilibrium,and the potential energy stored is 72.0 J.What compression (as measured from equilibrium) would result in 100 J being stored in this case?

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J.What is the spring constant in this case?

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

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 1 000 m.The parachute deploys,and she lands on the ground with a speed of 5.0 m/s.How much energy was lost to air friction during this jump?

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 baseball catcher puts on an exhibition by catching a 0.150-kg ball dropped from a helicopter at a height of 100 m above the catcher.If the catcher "gives" with the ball for a distance of 0.750 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

A 2 000-kg ore car rolls 50.0 m down a frictionless 10.0 $\degree$ 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 1.00 m?

A 20-N crate starting at rest slides down a rough 5.0-m long ramp,inclined at 25 $\degree$ with the horizontal.20 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?

Quiz 5: Energy

I drop a 60-g golf ball from 2.0 m high.It rebounds to 1.5 m.How much energy is lost?

A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37 °\degree° to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface,what is the force applied by the worker?

Adisa pulls a 40-N crate up a 5.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 37 °\degree° to the horizontal and there is a constant force of friction of 10 N between the crate and the surface,what is the net change in potential energy of the crate?

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 8.00 cm to 16.0 cm?

A girl and her bicycle have a total mass of 40 kg.At the top of the hill her speed is 5.0 m/s.The hill is 10 m high and 100 m long.If the force of friction as she rides down the hill is 20 N,what is her speed at the bottom?

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 speed received in the above situation?

A bobsled makes a run down an ice track starting at 150 m vertical distance up the hill.If there is no friction,what is the velocity at the bottom of the hill?

A pile driver drives a post into the ground.The mass of the pile driver is 2 500 kg and it is dropped through a height of 8.0 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?

An amount of work equal to 1.5 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 15-g marble?

A girl and her bicycle have a total mass of 40.0 kg.At the top of the hill her speed is 5.0 m/s,and her speed doubles as she rides down the hill.The hill is 10.0 m high and 100 m long.How much kinetic energy and potential energy is lost to friction?

A Hooke's law spring is compressed 12.0 cm from equilibrium,and the potential energy stored is 72.0 J.What compression (as measured from equilibrium) would result in 100 J being stored in this case?

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J.What is the spring constant in this case?

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

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 1 000 m.The parachute deploys,and she lands on the ground with a speed of 5.0 m/s.How much energy was lost to air friction during this jump?

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 baseball catcher puts on an exhibition by catching a 0.150-kg ball dropped from a helicopter at a height of 100 m above the catcher.If the catcher "gives" with the ball for a distance of 0.750 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s2)

A 2 000-kg ore car rolls 50.0 m down a frictionless 10.0 °\degree° 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 1.00 m?

A 20-N crate starting at rest slides down a rough 5.0-m long ramp,inclined at 25 °\degree° with the horizontal.20 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?

A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37 $\degree$ to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface,what is the force applied by the worker?

Adisa pulls a 40-N crate up a 5.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 37 $\degree$ to the horizontal and there is a constant force of friction of 10 N between the crate and the surface,what is the net change in potential energy of the crate?

A pile driver drives a post into the ground.The mass of the pile driver is 2 500 kg and it is dropped through a height of 8.0 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?

A baseball catcher puts on an exhibition by catching a 0.150-kg ball dropped from a helicopter at a height of 100 m above the catcher.If the catcher "gives" with the ball for a distance of 0.750 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

A 2 000-kg ore car rolls 50.0 m down a frictionless 10.0 $\degree$ 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 1.00 m?

A 20-N crate starting at rest slides down a rough 5.0-m long ramp,inclined at 25 $\degree$ with the horizontal.20 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?