A 280 kg piano slides 4.2 m down a 30° incline (with respect to the horizontal d
ID: 1346528 • Letter: A
Question
A 280 kg piano slides 4.2 m down a 30° incline (with respect to the horizontal direction) 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 work done by the normal force on the piano?_________J.
(f) What is the net work done on the piano?______J.
Thanks!
Explanation / Answer
a) Here, Fnet = 0
F_applied + mue_k*N - m*g*sin(30) = 0
F_applied = m*g*sin(30) - mue_k*N
= m*g*sin(30) - mue_k*m*g*cos(30)
= 280*9.8*sin(30) - 0.4*280*9.8*cos(30)
= 421.5 N
b) Wordone by man on piano = F_applied*d*cos(180)
= 421.5*4.2*(-1)
= -1770.3 J
c) work done by the friction force. = mue_k*N*d*cos(180)
= 0.4*280*9.8*cos(30)*4.2*(-1)
= -3992.3 J
d) work done by the force of gravity = m*g**sin(30)*d
= 280*9.8*4.2*sin(30)
= 5762.4 J
e) Workdone by normal force = N*d*cos(90)
= 0
f) Net workdone = sum of above all works
= 0
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