Question 1

A baseball catcher puts on an exhibition by catching a 0.170-kg ball dropped from a helicopter at a height of 110 m above the catcher.If the catcher "gives" with the ball for a distance of 0.695 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

Accepted Answer

First, we need to find the velocity of the ball just before it hits the mitt. Using the kinematic equation:

v^2 = u^2 + 2as

where v = final velocity (0 m/s, since the ball comes to rest), u = initial velocity (unknown), a = acceleration due to gravity (-9.80 m/s^2), and s = distance fallen (110 m).

Rearranging, we get:

u = sqrt(v^2 - 2as)

u = sqrt(-2(-9.80)(110)) = 47.0 m/s

Now, we can use the work-energy theorem to find the average force exerted on the mitt:

W = Fd = KEf - KEi

where W = work done by the mitt on the ball (negative since the force and displacement are in opposite directions), F = average force exerted on the mitt, d = distance the mitt "gives" (0.695 m), KEf = final kinetic energy of the ball (0 J), and KEi = initial kinetic energy of the ball (1/2 mv^2).

Rearranging, we get:

F = (KEi - KEf) / d = (1/2 mv^2) / d

F = (1/2)(0.170 kg)(47.0 m/s)^2 / 0.695 m = 264 N

Therefore, the answer is C) 264 N.

Question 2

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 kinetic energy of the mass is proportional to the square of its speed. Therefore, the compression of the spring must be increased by a factor of 2^2 = 4 to double the speed of the mass.

Question 3

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 potential energy stored in the spring is converted to gravitational potential energy. For the first mass m, the energy is mgh. For the second mass 2m, the energy is 2mg(h/2) = mgh, so the height is h/2.

Question 4

A girl and her bicycle have a total mass of 42 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 25 N,what is her speed at the bottom?

Accepted Answer

Using the conservation of energy principle, the initial gravitational potential energy at the top of the hill is transformed into kinetic energy and work done against friction at the bottom of the hill.

Initial potential energy = mgh = 42 kg x 9.8 m/s^2 x 10 m = 4116 J

Final kinetic energy = (1/2)mv^2, where v is the unknown speed at the bottom
Final work done against friction = Fd = 25 N x 100 m = 2500 J

Since energy is conserved, we can equate the initial potential energy to the sum of kinetic energy and work done against friction:

Initial potential energy = Final kinetic energy + Work done against friction
4116 J = (1/2)mv^2 + 2500 J
Solving for v, we get:
v = sqrt[(4116 J - 2500 J) x 2 / 42 kg] = 10.1 m/s

Therefore, the girl's speed at the bottom of the hill is 10.1 m/s.

Question 5

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 6

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 6.00 cm to 15.00 cm?

Accepted Answer

The answer of By how much is the energy stored...

Question 7

An amount of work equal to 2.5 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 25-g marble?

Accepted Answer

The work done on the spring is converted into kinetic energy of the marble. Using the formula for kinetic energy (KE = 0.5 * m * v^2), where m = 0.025 kg and KE = 2.5 J, solving for v gives approximately 14 m/s.

Question 8

A parachutist of mass 52.0 kg jumps out of an airplane at a height of 1 080 m.The parachute deploys,and she lands on the ground with a speed of 6.0 m /s.How much energy was lost to air friction during this jump?

Accepted Answer

The potential energy at the start is mgh = 52.0 kg * 9.8 m/s² * 1080 m = 550,080 J. The kinetic energy when landing is 0.5 * 52.0 kg * (6.0 m/s)² = 936 J. The energy lost to air friction is the difference: 550,080 J - 936 J = 549,144 J, which rounds to 549,000 J.

Question 9

A girl and her bicycle have a total mass of 43.0 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 15.0 m high and 100 m long.How much kinetic energy and potential energy is lost to friction?

Accepted Answer

The initial kinetic energy (KE_initial) is (1/2) * 43 * (5)^2 = 537.5 J. The final kinetic energy (KE_final) is (1/2) * 43 * (10)^2 = 2150 J. The potential energy at the top (PE_initial) is 43 * 9.8 * 15 = 6,321 J. The total mechanical energy at the top is KE_initial + PE_initial = 6,858.5 J. The total mechanical energy at the bottom is KE_final = 2150 J. The energy lost to friction is 6,858.5 J - 2150 J = 4,708.5 J, which rounds to 4,709 J.

Question 10

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 11

A pile driver drives a post into the ground.The mass of the pile driver is 2,700 kg and it is dropped through a height of 8.0 m on each stroke.If the resisting force of the ground is 3.0 × 10⁶ N,how far is the post driven in on each stroke?

Accepted Answer

The potential energy of the pile driver is converted into work done against the resisting force. The potential energy is mgh = 2,700 kg * 9.8 m/s² * 8.0 m = 211,680 J. The work done is force * distance, so 3.0 × 10⁶ N * distance = 211,680 J. Solving for distance gives approximately 0.071 m or 7.1 cm.

Question 12

A 2,000-kg ore car rolls 50.0 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 1.00 m?

Accepted Answer

The gravitational potential energy lost by the ore car is converted into the elastic potential energy stored in the spring. Using energy conservation: mgh = (1/2)kx^2. Calculate h using h = 50.0 m * sin(10.0°), then solve for k using k = 2mgh/x^2. The correct spring constant is approximately 347 kN/m.

Question 13

I drop a 65-g golf ball from 3.0 m high.It rebounds to 1.0 m.How much energy is lost?

Accepted Answer

Explanation:The initial potential energy of the ball is:PEi = mgh = (0.065 kg)(9.8 m/s^2)(3.0 m) = 1.96 JThe final potential energy of the ball is:PEf = mgh = (0.065 kg)(9.8 m/s^2)(1.0 m) = 0.637 JThe energy lost is:ΔE = PEi - PEf = 1.96 J - 0.637 J = 1.323 JTherefore, the correct answer is C) 1.32 J.

Question 14

A Hooke's law spring is compressed 15.0 cm from equilibrium,and the potential energy stored is 73.0 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 can be calculated using the formula:

PE = (1/2)kx^2

where k is the spring constant and x is the displacement from equilibrium.

Using the given information, we have:

73.0 J = (1/2)k(0.15 m)^2

Solving for k, we get:

k = 6513.3 N/m

To determine the compression required to store 100 J, we can rearrange the formula for potential energy and solve for x:

PE = (1/2)kx^2
100 J = (1/2)(6513.3 N/m)x^2
x^2 = 100 J / (1/2)(6513.3 N/m)
x^2 = 0.0306 m^2
x = 0.175 m

Therefore, the compression required to store 100 J is 0.175 m - 0.15 m = 0.025 m, which is equal to 2.5 cm. The total compression from equilibrium is therefore:

x_tot = 0.15 m + 0.025 m = 0.175 m = 17.6 cm

Therefore, the best choice is B.

Question 15

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 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 = 2PE / x^2. Substituting PE = 72.0 J and x = 0.12 m, we get k = 2(72.0) / (0.12)^2 = 10000 N/m.

Question 16

A 24-N crate starting at rest slides down a rough 4-m long ramp,inclined at 29° 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 gravitational potential energy lost is mgh = 24 N * 4 m * sin(29°). The work done by friction is 20 J. The kinetic energy at the bottom is the initial potential energy minus the work done by friction. Use KE = 0.5 * m * v^2 to solve for v.

Question 17

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 18

Adisa pulls a 40-N crate up a 6.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 38° 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, where m is mass, g is gravity, and h is height. The height (h) can be found using the sine of the incline angle: h = 6.0 m * sin(38°). The weight of the crate (40 N) acts as the force due to gravity. Thus, the change in potential energy is 40 N * 6.0 m * sin(38°) ≈ 148 J.

Question 19

Preston pushes a wheelbarrow weighing 490 N to the top of a 45.0-m ramp,inclined at 20.0° 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 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 work done by Preston is:$$W_p=F_pd=490	ext{ N}(45.0	ext{ m})\cos180\degree=-22050	ext{ J}$$The work done by friction is:$$W_f=F_fd=-80.0	ext{ N}(45.0	ext{ m})=-3600	ext{ J}$$The change in gravitational potential energy is:$$U_g=mgh=490	ext{ N}(45.0	ext{ m})\sin20.0\degree=17120	ext{ J}$$The total work done on the wheelbarrow is:$$W=W_p+W_f+U_g=-22050	ext{ J}-3600	ext{ J}+17120	ext{ J}=-8330	ext{ J}$$The kinetic energy at the bottom of the ramp is:$$K=\frac{1}{2}mv^2=W=-8330	ext{ J}$$

Question 20

A 40-N crate is pulled up a 4-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 10 N between the crate and the surface,what is the force applied by the worker?

Accepted Answer

The force applied by the worker must counteract both the gravitational component along the plane and the frictional force. The gravitational component is 40 N * sin(37°) ≈ 24 N. Adding the frictional force of 10 N, the total force required is 24 N + 10 N = 34 N.

Quiz 5: Energy

A baseball catcher puts on an exhibition by catching a 0.170-kg ball dropped from a helicopter at a height of 110 m above the catcher.If the catcher "gives" with the ball for a distance of 0.695 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

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 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 girl and her bicycle have a total mass of 42 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 25 N,what is her speed at the bottom?

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?

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 6.00 cm to 15.00 cm?

An amount of work equal to 2.5 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 25-g marble?

A parachutist of mass 52.0 kg jumps out of an airplane at a height of 1 080 m.The parachute deploys,and she lands on the ground with a speed of 6.0 m /s.How much energy was lost to air friction during this jump?

A girl and her bicycle have a total mass of 43.0 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 15.0 m high and 100 m long.How much kinetic energy and potential energy is lost to friction?

A pile driver drives a post into the ground.The mass of the pile driver is 2,700 kg and it is dropped through a height of 8.0 m on each stroke.If the resisting force of the ground is 3.0 × 10⁶ N,how far is the post driven in on each stroke?

A 2,000-kg ore car rolls 50.0 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 1.00 m?

I drop a 65-g golf ball from 3.0 m high.It rebounds to 1.0 m.How much energy is lost?

A Hooke's law spring is compressed 15.0 cm from equilibrium,and the potential energy stored is 73.0 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?

A 24-N crate starting at rest slides down a rough 4-m long ramp,inclined at 29° 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?

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

Adisa pulls a 40-N crate up a 6.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 38° 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 40-N crate is pulled up a 4-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 10 N between the crate and the surface,what is the force applied by the worker?

Quiz 5: Energy

A baseball catcher puts on an exhibition by catching a 0.170-kg ball dropped from a helicopter at a height of 110 m above the catcher.If the catcher "gives" with the ball for a distance of 0.695 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s2)

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 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 girl and her bicycle have a total mass of 42 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 25 N,what is her speed at the bottom?

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?

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 6.00 cm to 15.00 cm?

An amount of work equal to 2.5 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 25-g marble?

A parachutist of mass 52.0 kg jumps out of an airplane at a height of 1 080 m.The parachute deploys,and she lands on the ground with a speed of 6.0 m /s.How much energy was lost to air friction during this jump?

A girl and her bicycle have a total mass of 43.0 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 15.0 m high and 100 m long.How much kinetic energy and potential energy is lost to friction?

A pile driver drives a post into the ground.The mass of the pile driver is 2,700 kg and it is dropped through a height of 8.0 m on each stroke.If the resisting force of the ground is 3.0 × 106 N,how far is the post driven in on each stroke?

A 2,000-kg ore car rolls 50.0 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 1.00 m?

I drop a 65-g golf ball from 3.0 m high.It rebounds to 1.0 m.How much energy is lost?

A Hooke's law spring is compressed 15.0 cm from equilibrium,and the potential energy stored is 73.0 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?

A 24-N crate starting at rest slides down a rough 4-m long ramp,inclined at 29° 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?

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

Adisa pulls a 40-N crate up a 6.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 38° 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 40-N crate is pulled up a 4-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 10 N between the crate and the surface,what is the force applied by the worker?

A baseball catcher puts on an exhibition by catching a 0.170-kg ball dropped from a helicopter at a height of 110 m above the catcher.If the catcher "gives" with the ball for a distance of 0.695 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

A pile driver drives a post into the ground.The mass of the pile driver is 2,700 kg and it is dropped through a height of 8.0 m on each stroke.If the resisting force of the ground is 3.0 × 10⁶ N,how far is the post driven in on each stroke?