Conservation of energy: Block pulled up frictionless plane

[nickclarson]

Homework Statement

A block of mass 3.16 kg, starting from rest, is pulled up a frictionless inclined plane that makes an angle θ = 27.1° with the horizontal by a string parallel to the plane. The tension in the string is 45.8 N. After traveling a distance 2.12 m, the speed of the block is 3.37 m/s. Find the work done by the tension force.

Homework Equations

$$U=mgh$$
$$W = KE_{f} - KE_{i}$$

The Attempt at a Solution

I came up with:

$$KE_{i} + U_{i} = KE_{f} + U_{f}$$
$$0 + 0 = KE_{f} - U_{f}$$

so since I am looking to find the final kinetic energy which is equal to work...

$$KE_{f} = -U_{f} = -mgh$$

but positive because W is positive.

I got the wrong answer though and I could be way off, could somebody point me on the right track?

Thanks,
Nick

 

[anathema]

Dear Nick,

Thank you for sharing your attempt at solving this problem. It seems like you are on the right track, but there are a few things that may need clarification.

First, it is important to note that the work done by the tension force in this scenario is actually equal to the change in the kinetic energy of the block. This is because the only force acting on the block is the tension force, which is parallel to the displacement of the block. Therefore, the work done is given by the equation W = KE_{f} - KE_{i}, as you mentioned.

However, in your attempt at a solution, you set up the equation KE_{i} + U_{i} = KE_{f} + U_{f}. This equation is actually correct, but it is not necessary to use it in this scenario since the initial potential energy (U_{i}) is equal to zero, and the final potential energy (U_{f}) is also equal to zero because the block ends at the same height as it starts. Therefore, the equation simplifies to KE_{i} = KE_{f}.

Next, you mentioned that you got the wrong answer, but you did not provide the answer you got. It would be helpful to know the actual numbers you calculated to determine where the error may be.

To solve this problem, you can use the equation W = KE_{f} - KE_{i} and substitute in the given values for the final kinetic energy and the initial kinetic energy (which is zero since the block starts from rest). The final kinetic energy can be calculated using the equation KE_{f} = 1/2 * m * v^2, where m is the mass of the block and v is the final velocity. After substituting in the values, you should get the correct answer for the work done by the tension force.

I hope this helps. Keep up the good work!

 

[Moetasim]

Your attempt at a solution is partially correct. You correctly used the conservation of energy equation, but you made a mistake in your calculation for the final potential energy.

The correct calculation for the final potential energy is Uf = mghf, where hf is the final height of the block. Since the block is pulled up the inclined plane, the height will change from 0 to h = 2.12sin(27.1°) = 0.99 m.

Therefore, Uf = (3.16 kg)(9.8 m/s^2)(0.99 m) = 30.99 J.

Using the conservation of energy equation, we can then calculate the work done by the tension force:

W = KEf - KEi = (1/2)(3.16 kg)(3.37 m/s)^2 - 0 = 16.94 J.

Since there is no friction, all the work done by the tension force is converted into kinetic energy. Therefore, the work done by the tension force is 16.94 J.