Calculating Work Done by Friction?

[BrainMan]

Homework Statement

A 3-kg block is moved up a 37° incline under the action of a constant horizontal force of 40 N. The coefficient of kinetic friction is .1, and the block is displaced 2 m up the incline. Calculate (a) the work done by the 40-N force, (b) the work done by gravity, (c) the work done by friction, and (d) the change in kinetic energy of the block.

Homework Equations

The Potential and Kinetic Energy equations

The Attempt at a Solution

I managed to get all the parts of this question right except part (c). What I did was take the weight, which I found to be 29.4, and do 29.4 cos 37° to find the normal force. I got 23.48 for the normal force. Then I multiplied that by .1 to get the force of friction. Then I took the force of friction and multiplied it by the displacement to find the work done by friction. I got -4.7 J and the answer was -9.51 J.

 

[SammyS]

Homework Statement

A 3-kg block is moved up a 37° incline under the action of a constant horizontal force of 40 N. The coefficient of kinetic friction is .1, and the block is displaced 2 m up the incline. Calculate (a) the work done by the 40-N force, (b) the work done by gravity, (c) the work done by friction, and (d) the change in kinetic energy of the block.

Homework Equations

The Potential and Kinetic Energy equations

The Attempt at a Solution

I managed to get all the parts of this question right except part (c). What I did was take the weight, which I found to be 29.4, and do 29.4 cos 37° to find the normal force. I got 23.48 for the normal force. Then I multiplied that by .1 to get the force of friction. Then I took the force of friction and multiplied it by the displacement to find the work done by friction. I got -4.7 J and the answer was -9.51 J.

Draw a free body diagram for the block.

The horizontal force (actually the reaction to it) contributes to the normal force.

 

[Doc Al]

I managed to get all the parts of this question right except part (c). What I did was take the weight, which I found to be 29.4, and do 29.4 cos 37° to find the normal force. I got 23.48 for the normal force.

That would be fine except for that 40 N horizontal force. That force has a component normal to the surface, so it adds to the normal force on the block.

(SammyS beat me to it!)

 

[BrainMan]

OK I get it now. Thanks!

 

[HalfLostAndSearching]

I would like to commend you for your attempt at solving this problem and getting most of the parts correct. However, there are a few things that could be improved in your approach to calculating the work done by friction.

Firstly, in order to calculate the work done by friction, we need to use the equation W = Fd, where W is the work done, F is the force of friction, and d is the displacement. In your attempt, you correctly found the force of friction to be 2.348 N (23.48 N * 0.1), but you did not multiply it by the displacement of 2 m. Instead, you used the normal force to calculate the work done by friction, which is not correct.

Secondly, it is important to note that the work done by friction is always negative, as it acts in the opposite direction of the displacement. In your calculation, you obtained a positive value of 4.7 J, which implies that the work done by friction is in the same direction as the displacement, which is incorrect.

To obtain the correct value for the work done by friction, we can use the equation W = μN*d, where μ is the coefficient of kinetic friction, N is the normal force, and d is the displacement. Therefore, the work done by friction in this case would be -9.51 J, as you correctly mentioned.

In conclusion, as a scientist, I would suggest reviewing the equations and their applications to ensure that all calculations are correct and consistent with the given information. Keep up the good work!