A mover pushes a 260 kg piano on wheels up a ramp 7 meters long onto a stage 1.75 meters above the ground. The mover pushes the piano with a force of 680 N and the coefficient of friction is 0.0099.

Question

a) How much work does the mover do (concentrate on just the mover)?

b) What work is done on the piano by the ramp?

c) What is the speed of the piano at the top of the ramp?

d) How much change in KE at the top of the ramp if the piano started from rest?

e) How much net work would be done?

Answer 1

(a)

If we concentrate just on the mover. We only take into account the force it applies onto the piano and gravitation work.

Wmover = F * L = 680 * 7 - 260*9.81*1.75 = 296.45 J

(b)

Need to find the work of friction from the ramp to piano

sinθ = 1.75/7 = 14.477 degrees

Fn = mgcosθ = 260*9.81cos14.477 = 2550.6 N

Fk = μ_k * Fn = 0.0099 * 2550.6 = 25.25 N

Wfriction = Wramp = - Fk * L = -25.25 * 1.75 = - 44.1875 J

Work is negative of friction since it opposes line of motion

c)

V1 = 0

V2 = ?

Sum of Work = Delta KE

296.45-44.19 = (1/2) * m * (V2^2-V1^2)

(2/260) * (296.45-44.1875) = V2^2

V2^2 = 1.940

V2 = 1.39 m/s

d) ΔKE = (1/2) * m * (V2^2-V1^2)

V1 = 0

V2 = 1.39

ΔKE = (1/2) * 260 * 1.39^2 = 252.2 J

e) Total Work = 296.45-44.19 = 252.26 J

Answer 2