Question

A 290 kg piano slides 4.3 m down a 30 degree 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)?

Answer 1

Part A. Given that man keeps piano from accelerating by pushing back the piano on the incline, So net force exerted by the man will be equal to total downward force along the incline. So Using FBD:

Force along the incline will be:

F_net = W*sin - Ff - F

Since acceleration is zero, So F_net = 0 & F = Force exerted by man

which means

W*sin - Ff - F = 0

F = W*sin - Ff

W = Weight of piano = m*g

Ff = friction force applied = *N = *m*g*cos

F_net = m*g*sin - *m*g*cos

Using given values:

F_net = 290*9.8*sin 30 deg - 0.40*290*9.8*cos 30 deg

F_net = 436.5 N

Part B.

Work-done is given by:

W = F*d*cos

F = Force applied by man = 436.5 N upward

d = displacement of piano = 4.3 m downward

= angle between Force and displacement = 180 deg

W = 436.5*4.3*cos 180 deg = -1876.95 J

W = -1877 J

Part C.

Friction force on piano is in upward direction, which is given by:

Ff = friction force applied = *N = *m*g*cos

Ff = 0.40*290*9.8*cos 30 deg = 984.5 N

d = displacement of piano = 4.3 m downward

= angle between friction force and displacement = 180 deg

W1 = 984.5*4.3*cos 180 deg = -4233.35 J

W1 = -4233 J

Part D.

Force of gravity on the piano is downward which is given by:

Fg = W*sin = 290*9.81*sin 30 deg = 1421 N

d = displacement of piano = 4.3 m downward

= angle between friction gravity force and displacement = 0 deg

W2 = 1421*4.3*cos 0 deg = 6110.3 J

W2 = 6110 J

Part E.

Net work-done on piano will be:

W_net = W + W1 + W2

W_net = -1877 - 4233 + 6110

W_net = 0 J

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