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A 295 kg piano slides 4.5 m down a 30° incline and is kept from accelerating by

ID: 2022811 • Letter: A

Question

A 295 kg piano slides 4.5 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40. I have tried for over 4 hours to get this problem right, but no luck. I am willing to give MAX Karma points to anyone who can provide ALL solutions and ALL answers. Thanks. See Link

http://i808.photobucket.com/albums/zz5/buymovieposters/page7-1.jpg

(a) Calculate the force exerted by the man.
_445_ N (Correct)
(b) Calculate the work done by the man on the piano.
__?__J
(c) Calculate the work done by the friction force.
__?__ J
(d) What is the work done by the force of gravity?
__?__ J
(e) What is the net work done on the piano?
__?__ J






Explanation / Answer

mass m = 295 kg distance S= 4.5 m angle = 30 degrees The effective coefficient of kinetic friction is = 0.40 force exerted by the man F = 445 N (b) the work done by the man on the piano W = FS                                                                       = 2002.5 J (c) the work done by the friction force W ' = -mg cos * S                                                                  = -4506.62 J
(d) the work done by the force of gravity W " = -mg sin * S                                                                       = - 6504.75 J (e) the net work done on the piano = W + W ' + W "                                                      = -9008.87 J
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