A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angl

Question

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 27.0 above the horizontal by a force F  of magnitude 146 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.270. The suitcase travels a distance 4.00 m along the ramp.

1-Calculate the work done on the suitcase by the gravitational force.

2-Calculate the work done on the suitcase by the friction force.

3-Calculate the total work done on the suitcase.

4-If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 4.00 malong the ramp?

Explanation / Answer

N = mg cos theta = 19.6 x 9.8 x .891= 171.14 N

friction force (f) = mu N = .27 x 171.14 = 46.2 N

Fg = mg sin theta = 19.6 x 9.8 x .45 = 87.2 N

Total net force = F - (Fg + f) = 146 - (87.2 + 46.2) = 12.6 N

(a) Wg = Fg x d = 87.2 x 4 = 348.8 J

(b) Wf = f x d = 46.2 x 4 = 184.8 J

(c) Wt = net F x d = 12.6 x 4 = 50.4 J

(d) acceleration = net F / m = 12.6 / 19.6 = .642 m/s^2

V^2 = u^2 + 2aS

V^2 = 0 + (2 x .642 x 4)

V = 2.26 m/s

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