What is the work done by kinetic friction on a block sliding down a ramp?

[pdrjuarez]

Homework Statement

A 25.0 kg block slides down a 5.00 m ramp that is elevated at 36.0 degrees. The kinetic friction coeff. is 0.220.
-What is the work done by friction as the block slides down the ramp?

Ok, so as far as I know, work is achieved only if you apply a force on something and move that something in the same direction as the force. Here, friction is applying a force of 43N (I calculated it), but it's not achieving motion in the same direction. Therefore, work done by friction is 0.

Am I right?

 

[aba3]

Work done is not 0, its simply negative.
You are right that W = Fd. But friction does apply a force, and the object does move, just in the opposite direction. So the work is F*(-d)

 

[RoyalCat]

Homework Statement

A 25.0 kg block slides down a 5.00 m ramp that is elevated at 36.0 degrees. The kinetic friction coeff. is 0.220.
-What is the work done by friction as the block slides down the ramp?

Ok, so as far as I know, work is achieved only if you apply a force on something and move that something in the same direction as the force. Here, friction is applying a force of 43N (I calculated it), but it's not achieving motion in the same direction. Therefore, work done by friction is 0.

Am I right?

Nope. :) Positive work is produced when the force and displacement are parallel.

Take a look at the vector expression for work:

W=\vec F\cdot \vec d

W=Fd\cos{\theta} where \theta is the angle between the force and displacement vectors.

This way, I hope you can see that work can be either positive, negative, or zero. All depending on \cos{\theta}

 

[Senjai]

Review your notes

Straight out of a physics textbook, Work done on a particle by a constant force (constant magnitude and direction) is defined to be the product of the magnitude of its displacement (d) times the component of the force parallel to its displacement.

In equation form, W=F_{net}d \cdot cos\theta

Remember that it says Fnet, not F. Remember also, there is only a net force on the object if a\neq0 So you can only do work if there is a net force on the object. The exception is if the question is asking you to determine the work done by a specific force, other than the net force.

In your question the force of friction IS your net force. So assuming your calculation for the force of friction is correct (i won't check it) Fnet = 43N. Remember, Friction opposes the direction of motion ALWAYS. What way is your object moving? In which direction does the force of friction act?

remember that when theta is 180 degrees (opposite to the direction of motion) cos 180 degrees equals -1. Which would imply the work done by friction is always NEGATIVE...

Try again and let me know :)