Work done by friction

[Hughey85]

Question:

A 1-kg ball starting at h = 8.6 meters slides down a smooth surface where it encounters a rough surface and is brought to rest at B, a distance 15.8 meters away. To the nearest joule what is the work done by friction?

Can you pls. help with this question? Do you need to find the potential energy and then work from there? potential E = mgy so you could find that... U = (1 kg) * (9.81 m/s) * (8.6 m) = 84.366

but how would I work it from there?

 

[stunner5000pt]

i don't understand is the surface inclined or what

what is the 15.8m supposed to be the horizontal component (if it is inclined plane)?

 

[Hughey85]

The surface is a slope that evens out at the end...like a waterslide...by a smooth surface, I believe that means "frictionless"...the 15.8 meters represents the distance when you hit the rough surface to when you stop. The rough surface starts at the "bottom of the slide" and runs along the x-axis.

I hope this isn't confusing, the picture isn't, but I can't seem to copy it over.

 

[stunner5000pt]

from what you just explained

dU = dK from the top of the slide to the bottom all the potential is converted to kinetic energy

mgh = 0.5 m v^2

v1 = root (2gh)

now dK = Mu Fn d

where Mu is the coefficient, Fn is the normal force, and d is the dsitnace it travelled