Change in Kenetic Energy

Question

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. If the piano started at rest, how much change in KE is there when it reaches the top of the ramp?

(Assuming that gravity is 10 m/s)

Answer 1

Anthony,

The question asking for kinetic energy at the end of motion that started from rest means that the piano must go up the ramp with acceleration.

For the piano move up with an acceleration a, the force F_mov applied by the mover to the piano must by an amount ma be OVER the sum of two forces working against the pull :

1. The sliding force F_sl due to the ramp's incline at an angle θ,

1. F_sl=mgsinθ (1)

2. The force of friction F_fr - a product of the normal force N=mgcosθ with which the piano would press on the ramp

and the coefficient of kinetic friction μ:

1. F_fr=μmgcosθ (2)

The net force acting on the piano would be equal to

F_net=F_mov- (F_sl+F_fr)=F_mov-(mgsinθ + μmgcosθ) (3)

The change of kinetic energy of a body acted on by a force F_net along a path of length L equals to the work of that force

ΔKE=F_metL=L[F_mov-(mgsinθ + μmgcosθ)] ,(4)

Now, let's express sinθ and cosθ through the length L and the elevation H of the ramp:

sinθ=H/L. ; cosθ,=Sqrt(L²-H²)/L. (5)

Substituting (5) into (4) we'd get

ΔKE=F_metL=

L[F_mov-(mgsinθ + μmgcosθ)=LF_mov - mg(H +μSqrt(L²-H²)](6)

Using given data F_mov=680N; µL=7m; H=1.75m; g=10m/s³; m=260 kg; µ=0.0099, (7)

we'd get the result (in units of J)

ΔKE=7x680-260x10[1.75+0.0099xSqrt(7²-1.75²)=170J (if we:d keep 2 sig.figs)

Note: To be consistent in applying the rules of sig.figs the result should be rounded to only ONE sig. fig. in (6) as L=7m is given with only 1sog.fig.

should be reported as 200 J

Dr.Ariel.B