- #1
reaperkid
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Homework Statement
Alrighty, I have most of this set but I'm getting the wrong answer and I'm not sure why.
A block of mass 18.0 kg is sliding down an 6.0 metre long ramp inclined at 56.0 deg. to the horizontal. If the coefficient of kinetic friction between the ramp and the block is 0.48, how much work is done by friction as the block moves from the top to the bottom of the ramp ?
m = 18kg
d = 6 m
Theta = 56 degrees
mu = .48
Homework Equations
W = Fd
F = ma
mu = fk / fn
The Attempt at a Solution
First I calculated work done by gravity. . .
W(g) = Fd = (18)(9.8)(6)(cos 56) = 591.85 J
Then I attempted to calculate the work done by friction. . .
W(f) = (.48*18*9.8*6*sin56) = 421.178 J
I tried that as a negative number as well, since the distance is negative.
So, clearly I'm missing something.
Any help would be greatly appreciated. Thanks!