A 255 kg piano slides 4.0 m down a 30° incline and is kept from accelerating by a man who is pushing back on?
...it parallel to the incline. The effective coefficient of kinetic friction is 0.40.
(a) Calculate the force exerted by the man.
N
(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
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Mass of piano M= 255kg, Angle of inclination or A=30°, Coefficient of friction or μ= 0.4
a) Force exerted by man or F= mgsinA-μmgcosA=255 x10 x sin30°- 0.4 x 255cos30°= 1275-883.34=391.6N
b) Work done by man=396.6 x 4(slides 4m)= 1586.4J
C)Work done by Friction force= μmgcosA x 4(as it slides 4m)= 4 x 883.34= 3533.36J
d)Work done by gravity = mg x 4sinA= 255 x10 x 4 x sin30= 5100J
e)Add those work for net work done as its a scalar quantity!