Help with total work done by friction problem

1. Oct 22, 2009

debrickshaw

1. The problem statement, all variables and given/known data
A Block of mass (m) slides down a plank of length (s) from a height (h) down to the ground. The friction between the plank and the block is uk. Derive and expression for the total work done by the force of friction done by force of friction on the block

2. Relevant equations
W=Fd
F=W/d
d=W/F
Wtotal=1/2 mv^2

3. The attempt at a solution
at top PE = M=mgh
honestly ive been staring at this damn problem for a half an hour and thats all ive got, any help would be very appreciated

2. Oct 22, 2009

cepheid

Staff Emeritus
Work equals force times distance. Can you figure out what the magnitude of the friction force on the block is?

3. Oct 22, 2009

debrickshaw

ok, so mg(mk) is the magnitude of the friction force
so w = mg(mk) * s?

4. Oct 22, 2009

debrickshaw

i mean mg (uk)

5. Oct 22, 2009

cepheid

Staff Emeritus
Careful. The friction force is equal to the normal force times uk. Is the normal force on the block equal to its weight in this situation?

6. Oct 22, 2009

debrickshaw

ok, so F of friction is equal to sin (theta) x uk x mg?

7. Oct 22, 2009

debrickshaw

ok so this is what i've worked out so far

W= F x D
D= -SF=uk x m x g

but i dont think that's right, i cant figure out how to incorporate the height, nor do i know how to get the normal force

8. Oct 22, 2009

cepheid

Staff Emeritus
Draw a diagram. Resolve the block's weight into two components, one in the direction parallel to the ramp, and one in the direction perpendicular to the ramp. The normal force is equal in magnitude and opposite in direction to this perpendicular component of the block's weight (we know this because there is no acceleration in the direction perpendicular to the ramp).

Once you have your diagram, you will be able to see that the angle between the total weight vector and its perpendicular component is the same as the angle of inclination of the ramp.

Hint: In your second last post, sin(theta) is not correct.