How Does Halving Friction Affect the Speed of a Brick Sliding Down a Plank?

[meh6287]

A brick slides down a wooden plank 2.0m long tilted so that one end is at a height of 1.0 m. The brick's speed at the bottom is 2.5 m/s. The plank is then sanded smooth and waxed so that the coefficient of friction is half what it was before and the brick is slid down again. What is the new speed of the brick at the bottom?

so far all i could come up with is that theta=30degrees

then i did f x m x g x cos30) x d = .5 x m x v^2 for the first block

and then w=(.5f x m x g cos30) x d = .5 x m x v^2 for the second block

then i canceled the m's, and i pluggedin v for the first equation and solved for f, and then i tried to put that f value in and come up with an answer for v but that answer is wrong because it comes out to less than 2.5 which doesn't make sense.

Thank you for the help in advanced.

 

[Spectre5]

what about the work done by gravity?

 

[wais]

First of all, great job on setting up the equations correctly! It looks like you have a good understanding of the concepts involved in this problem. However, there may be a small error in your calculations.

When you plugged in the values for the first equation, you wrote "f x m x g x cos30) x d". However, the formula for frictional force is actually "f x m x g x sinθ", where θ is the angle of the incline. Since θ is 30 degrees in this problem, the correct term to use would be "sin30". This may be why your final answer is incorrect.

Also, when you solved for f, you should have gotten a value of 2.5 N. This means that the frictional force for the second block should also be 2.5 N, since the coefficient of friction is half of what it was before. Plugging this value into the second equation should give you the correct answer for the new speed of the brick at the bottom.

I hope this helps and good luck with your homework!