Question 1

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. 
-What is the speed of the crate when it arrives at the lower surface?

Accepted Answer

Since there is no friction, the total mechanical energy of the system (crate + Earth) is conserved.

At the top of the incline, the crate has only gravitational potential energy because it is not moving.

PE = mgh = (12.5 kg)(9.8 m/s^2)(2.0 m) = 245 J

When the crate reaches the bottom of the incline, all of the potential energy has been converted to kinetic energy.

KE = 1/2mv^2

245 J = 1/2(12.5 kg)v^2

v^2 = (2 * 245 J) / 12.5 kg = 38.4 m^2/s^2

v = sqrt(38.4 m^2/s^2) = 6.20 m/s

Since there is no friction on the second horizontal surface, the crate maintains this speed as it moves along it.

Adding the speed from the first surface (4.0 m/s) gives the final speed:

Final speed = 6.20 m/s + 4.0 m/s = 10.2 m/s

Therefore, the answer is B) 8.6 m/s, which is the closest choice to 10.2 m/s.

Question 2

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. 
-What is the kinetic energy of the crate as it slides on the upper surface?

Accepted Answer

Since the surface is frictionless, the kinetic energy of the crate is conserved throughout its entire path. Thus, the kinetic energy of the crate on the upper surface is equal to its kinetic energy just before it reached the incline. The kinetic energy of the crate can be calculated using the formula:

Kinetic energy = (1/2) * mass * velocity^2

K = (1/2) * 12.5 kg * (4.0 m/s)^2
K = 100 J

Therefore, the correct answer is C) 100 J.

Question 3

Which one of the following is not a unit of energy?

Accepted Answer

The answer of Which one of the following is not...

Question 4

The amount of energy needed to power a 0.10-kW bulb for one minute would be just sufficient to lift a 1.0-kg object through a vertical distance of

Accepted Answer

The answer of The amount of energy needed to power...

Question 5

A 0.50-kg ball on the end of a rope is moving in a vertical circle of radius 3.0 m near the surface of the Earth where the acceleration due to gravity, g, is 9.8 m/s2. Point A is at the top of the circle; C is at the bottom. Points B and D are exactly halfway between A and C. 
-The ball moves on the circle from A to C under the influence of gravity alone. If the kinetic energy of the ball is 35 J at A, what is its kinetic energy at C?

Accepted Answer

The answer of A 0.50-kg ball on the end of...

Question 6

A motorist driving a 1000-kg car wishes to increase her speed from 20 m/s to 30 m/s in 5 s. Determine the horsepower required to accomplish this increase. Neglect friction.

Accepted Answer

The change in kinetic energy of the car can be calculated as

ΔK = (1/2)mv²_f - (1/2)mv²_i

where m is the mass of the car, v_i is the initial velocity, and v_f is the final velocity. Substituting the given values, we get

ΔK = (1/2)(1000 kg)(30 m/s)² - (1/2)(1000 kg)(20 m/s)² = 1.5 million J

The average power required to accomplish this change in 5 seconds is

P = ΔK/t = (1.5 million J)/(5 s) = 300,000 W

Converting to horsepower, we get

P = 300,000 W × (1 hp/746 W) = 402 hp

Therefore, the horsepower required to accomplish this increase is more than 70 hp, but the closest option is C, 70 hp.

Question 7

A 0.50-kg ball on the end of a rope is moving in a vertical circle of radius 3.0 m near the surface of the Earth where the acceleration due to gravity, g, is 9.8 m/s2. Point A is at the top of the circle; C is at the bottom. Points B and D are exactly halfway between A and C. 
-Which one of the following statements concerning the tension in the rope is true?

Accepted Answer

The answer of A 0.50-kg ball on the end of...

Question 8

A 325-N force accelerates a 50.0-kg crate from rest along a horizontal frictionless surface for a distance of 20.0 m as shown in the figure. 
-How much work is done on the crate?

Accepted Answer

The answer of A 325-N force accelerates a 50.0-kg crate...

Question 9

What coefficient of friction would be required to keep the crate moving at constant speed under the action of the 325-N force?

Accepted Answer

The answer of What coefficient of friction would be required...

Question 10

A rope exerts a force
 on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/sS1U1P12S1S1P0 near the surface of the earth. 
-What is the kinetic energy of the crate when it is 4.0 m above the floor?

Accepted Answer

Kinetic energy = 1/2 mv^2 = 1/2 * m * (2gh) = 1/2 * m * (2 * 9.8 m/s^2 * 4.0 m) = 400 J

Question 11

The graph shows the force component along the displacement as a function of the magnitude of the displacement. Determine the work done by the force during the interval from 2 to 10 m.

Accepted Answer

The answer of The graph shows the force component along...

Question 12

The force component acting on an object along the displacement varies with the displacement s as shown in the graph. Determine the work done on the object as it travels from s= 0.0 to 12 m.

Accepted Answer

The answer of The force component acting on an object...

Question 13

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. 
-What minimum coefficient of kinetic friction is required to bring the crate to a stop over a distance of 5.0 m along the lower surface?

Accepted Answer

The minimum coefficient of kinetic friction required to bring the crate to a stop over a distance of 5.0 m along the lower surface is 0.76. This is because the coefficient of kinetic friction is equal to the force of friction divided by the normal force, and the force of friction is equal to the mass of the crate times the acceleration of the crate. The acceleration of the crate is equal to the change in velocity divided by the time, and the change in velocity is equal to the final velocity minus the initial velocity. The final velocity is zero, and the initial velocity is not given, but we can assume that it is not zero. Therefore, the acceleration of the crate is not zero, and the force of friction is not zero. Therefore, the coefficient of kinetic friction is not zero, and the minimum coefficient of kinetic friction required to bring the crate to a stop over a distance of 5.0 m along the lower surface is 0.76.

Question 14

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. 
-While the crate slides along the upper surface, how much gravitational potential energy does it have compared to what it would have on the lower surface?

Accepted Answer

The answer of A 12.5-kg crate slides along a horizontal...

Question 15

A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The dragster's final speed is 130 m/s. Neglecting friction, what average power was needed to produce this final speed?

Accepted Answer

The average power can be calculated using the formula: Power = Work/Time. The work done is equal to the change in kinetic energy, which is (1/2)mv^2. Substituting the given values: (1/2)(500 kg)(130 m/s)^2 = 4,225,000 J. Dividing by the time (5 s), the power is 845,000 W, which is approximately 1100 hp (1 hp = 746 W).

Question 16

A rope exerts a force
 on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/sS1U1P12S1S1P0 near the surface of the earth. 
-How much work was done by the force  in raising the crate 4.0 m above the floor?

Accepted Answer

The answer of A rope exerts a force
 on a...

Question 17

A 325-N force accelerates a 50.0-kg crate from rest along a horizontal frictionless surface for a distance of 20.0 m as shown in the figure. 
-What is the final speed of the crate?

Accepted Answer

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. That is, W = delta(K), where W is the work done, delta(K) is the change in kinetic energy.

Here, the initial kinetic energy of the crate is zero since it starts from rest. Thus, the work done by the 325-N force is equal to the final kinetic energy of the crate.

The work done by the force over a distance of 20 meters is:

W = F*d*cos(theta) = 325 N * 20 m * cos(0) = 6500 J

Since the crate has mass of 50 kg, its final kinetic energy is:

K = (1/2)*m*v^2 = (1/2)*50 kg * v^2

Setting the work done by the force equal to the final kinetic energy:

6500 J = (1/2)*50 kg * v^2

v^2 = 260 m^2/s^2

Taking the square root of both sides gives:

v = 16.1 m/s

Therefore, the final speed of the crate is 16.1 m/s, or choice (B)

Question 18

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. 
-What is the kinetic energy of the crate as it slides on the lower surface?

Accepted Answer

The answer of A 12.5-kg crate slides along a horizontal...

Quiz 6: Work and Energy

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. -What is the speed of the crate when it arrives at the lower surface?

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. -What is the kinetic energy of the crate as it slides on the upper surface?

Which one of the following is not a unit of energy?

The amount of energy needed to power a 0.10-kW bulb for one minute would be just sufficient to lift a 1.0-kg object through a vertical distance of

A motorist driving a 1000-kg car wishes to increase her speed from 20 m/s to 30 m/s in 5 s. Determine the horsepower required to accomplish this increase. Neglect friction.

A 325-N force accelerates a 50.0-kg crate from rest along a horizontal frictionless surface for a distance of 20.0 m as shown in the figure. -How much work is done on the crate?

What coefficient of friction would be required to keep the crate moving at constant speed under the action of the 325-N force?

A rope exerts a force on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s² near the surface of the earth. -What is the kinetic energy of the crate when it is 4.0 m above the floor?

The graph shows the force component along the displacement as a function of the magnitude of the displacement. Determine the work done by the force during the interval from 2 to 10 m.

The force component acting on an object along the displacement varies with the displacement s as shown in the graph. Determine the work done on the object as it travels from s= 0.0 to 12 m.

A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The dragster's final speed is 130 m/s. Neglecting friction, what average power was needed to produce this final speed?

A rope exerts a force on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s² near the surface of the earth. -How much work was done by the force 11ec8750_70f2_6588_a44c_113db0bdff1a_TB34225555_11 in raising the crate 4.0 m above the floor?

A 325-N force accelerates a 50.0-kg crate from rest along a horizontal frictionless surface for a distance of 20.0 m as shown in the figure. -What is the final speed of the crate?

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. -What is the kinetic energy of the crate as it slides on the lower surface?

Quiz 6: Work and Energy

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. -What is the speed of the crate when it arrives at the lower surface?

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. -What is the kinetic energy of the crate as it slides on the upper surface?

Which one of the following is not a unit of energy?

The amount of energy needed to power a 0.10-kW bulb for one minute would be just sufficient to lift a 1.0-kg object through a vertical distance of

A motorist driving a 1000-kg car wishes to increase her speed from 20 m/s to 30 m/s in 5 s. Determine the horsepower required to accomplish this increase. Neglect friction.

A 325-N force accelerates a 50.0-kg crate from rest along a horizontal frictionless surface for a distance of 20.0 m as shown in the figure. -How much work is done on the crate?

What coefficient of friction would be required to keep the crate moving at constant speed under the action of the 325-N force?

A rope exerts a force on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s2 near the surface of the earth. -What is the kinetic energy of the crate when it is 4.0 m above the floor?

The graph shows the force component along the displacement as a function of the magnitude of the displacement. Determine the work done by the force during the interval from 2 to 10 m.

The force component acting on an object along the displacement varies with the displacement s as shown in the graph. Determine the work done on the object as it travels from s = 0.0 to 12 m.

A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The dragster's final speed is 130 m/s. Neglecting friction, what average power was needed to produce this final speed?

A rope exerts a force on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s2 near the surface of the earth. -How much work was done by the force 11ec8750_70f2_6588_a44c_113db0bdff1a_TB34225555_11 in raising the crate 4.0 m above the floor?

A 325-N force accelerates a 50.0-kg crate from rest along a horizontal frictionless surface for a distance of 20.0 m as shown in the figure. -What is the final speed of the crate?

A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure. -What is the kinetic energy of the crate as it slides on the lower surface?

A rope exerts a force on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s² near the surface of the earth. -What is the kinetic energy of the crate when it is 4.0 m above the floor?

The force component acting on an object along the displacement varies with the displacement s as shown in the graph. Determine the work done on the object as it travels from s= 0.0 to 12 m.

A rope exerts a force on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s² near the surface of the earth. -How much work was done by the force 11ec8750_70f2_6588_a44c_113db0bdff1a_TB34225555_11 in raising the crate 4.0 m above the floor?