Original problem: A worker pushed a 21 kg block 7.2 m along a level floor at con

Question

Original problem:
A worker pushed a 21 kg block 7.2 m along a level floor at constant speed with a force directed 38° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.23, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

I found the answers to be:
(a) 415 J
(b) 414 J

using the methods found:
http://answers.yahoo.com/question/index?qid=20091102074529AAiEhN3

My answers are correct but is there an easier way to found them for next time? Thanks!

Explanation / Answer

since block is moving with constant velocity,
horizontal force component is equal to frictional force
Fcos38=*(mg+Fsin38)

Fcos38=0.23*(21*9.8+Fsin38)

F(0.6464=47.334

F=73.23N

work done by worker,force=Fcos38*d

=73.23*c038*7.2

=415.483J

this work done will be converted to thermal energy.

therefore,the increase in thermal energy of the block-floor system=415.483J

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