View Full Version : Work done by frition on a crate!
cbeeson23
Sep20-09, 05:15 PM
A 70 kg crate starting at rest slides down a rough ramp 4.0 meters long, inclined at 30 degrees above horizontal. At the bottom, it is moving at 5.0 m/s. what was the work done by friction on the crate???
Answers=
A. -500 Joules
B. 500 Joules
C. 460 Joules
D. -460 Joules
E. 0.0 Joules
I got this answer wrong and i was wondering why... so here is my work. i got the height to be 2meters by using sin 30
Mgh - Wf = 1/2MV^2
70 * 9.81 * 2 - wf = .5 * 70 * 5.0^2
875 - wf = 1373.4
wf =1373.4-875
wf = 500 joules
so i got positive 500 joules. can any body tell me what the answer truly is and what i did wrong?? thank you alot!!
tiny-tim
Sep20-09, 06:17 PM
Hi cbeeson23! Welcome to PF! :smile:
(try using the X2 tag just above the Reply box :wink:)
A 70 kg crate starting at rest slides down a rough ramp 4.0 meters long, inclined at 30 degrees above horizontal. At the bottom, it is moving at 5.0 m/s. what was the work done by friction on the crate???
…
so i got positive 500 joules.
Work done = force "dot" displacement …
the friction is up, but the displacement is down. :wink:
cbeeson23
Sep20-09, 06:22 PM
so the answer then would be -500 joules?
should my equation have been mgh = 1/2mv - work
cbeeson23
Sep20-09, 06:54 PM
Or are you saying that the work done is actually 500 joules, but since friction is what goes up the ramp the work done by friction is actually - 500 joules?
cbeeson23
Sep20-09, 07:18 PM
can any1 explain this answer to me a little better
rl.bhat
Sep20-09, 07:38 PM
can any1 explain this answer to me a little better
Frictional force acts in the opposite direction of the displacement of the crate.
When the work is done on a body, its kinetic energy increases.
When the work is done by a body, its kinetic energy decreases.
When there is no friction, mgh = 1/2*mv^2
Here mgh > 1/2*mv^2
So due to friction kinetic energy of the crate decreases.
Hence work done by the friction = final energy - initial energy.
tiny-tim
Sep21-09, 02:45 AM
Hi cbeeson23! :smile:
(just got up :zzz: … )
If you're confused, then don't treat potential energy as part of the energy, treat it as part of the work done (PE is just another name for minus work done in a conservative field, such as gravity).
The formula to remember is ∆(kinetic energy) = total work done …
which in this case is mgh - Fd :smile:
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