- #1

ag3

- 1

- 0

- Homework Statement
- A person pulls a mass m (initially at rest) all the way to the top of an incline and then allows it to slide halfway down the incline before exerting enough tension force to stop the mass at this halfway point. If the coefficient of kinetic friction is mu, the length of the incline is d, the angle of inclination is theta, what is the total work done by the person?

- Relevant Equations
- Change in Mechanical Energy = Work Nonconservative

Kinetic Friction Force = mu mgcos(theta)

Gravitational Potential Energy = mgh

There are two nonconservative forces in this situation, the work done by the person and the work done by friction - they are the only sources of work that change the total mechanical energy of the mass-Earth system.

The initial energy (assuming gravitational potential energy is initially 0) is 0 and the final energy would only consist of gravitational potential energy, which is

(1/2)mgdsin(theta)

Thus,

(1/2)mgdsin(theta) = Work by person - (3/2) mu*mgdcos(theta)

so

Work by person = (1/2)mgdsin(theta) + (3/2) mu*mgdcos(theta)

I get how to solve the problem, but want more understanding on the general physics of the situation presented in the problem, so I have 2 questions:

1. If the person who did the work allowed the block to fall further down the incline, friction would have done more work on the block, but the change in gravitational potential energy, and thus total mechanical energy, would decrease. Do these differences compensate so that the work done by the person stays constant no matter what height the block falls to? If not, does the person's work increase or decrease as he/she allows the block to fall more of a distance until he stops it?

2. How does kinetic energy play into this situation? Let's say the person did not stop the block as it was going back down the ramp and that the problem asked to find the work done by the person as the block was halfway down and still moving (assuming the work done by friction does not make it stop moving entirely). Would this answer differ from the answer from the original problem and how so?

Thanks again. I know these questions are a little bit off from what the problem was asking, but I like asking these questions to gain a better intuition of the situations presented in physics problems.

The initial energy (assuming gravitational potential energy is initially 0) is 0 and the final energy would only consist of gravitational potential energy, which is

(1/2)mgdsin(theta)

Thus,

(1/2)mgdsin(theta) = Work by person - (3/2) mu*mgdcos(theta)

so

Work by person = (1/2)mgdsin(theta) + (3/2) mu*mgdcos(theta)

I get how to solve the problem, but want more understanding on the general physics of the situation presented in the problem, so I have 2 questions:

1. If the person who did the work allowed the block to fall further down the incline, friction would have done more work on the block, but the change in gravitational potential energy, and thus total mechanical energy, would decrease. Do these differences compensate so that the work done by the person stays constant no matter what height the block falls to? If not, does the person's work increase or decrease as he/she allows the block to fall more of a distance until he stops it?

2. How does kinetic energy play into this situation? Let's say the person did not stop the block as it was going back down the ramp and that the problem asked to find the work done by the person as the block was halfway down and still moving (assuming the work done by friction does not make it stop moving entirely). Would this answer differ from the answer from the original problem and how so?

Thanks again. I know these questions are a little bit off from what the problem was asking, but I like asking these questions to gain a better intuition of the situations presented in physics problems.