A 280 kg piano slides 4.2 m down a 30° incline (with respect to the horizontal d

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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