Question

A luggage handler pulls a 18.0-kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 169 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.250. The suitcase travels 3.80 m along the ramp.

a)Calculate the work done on the suitcase by F⃗ . Express your answer with the appropriate units.

b)Calculate the work done on the suitcase by the gravitational force.

c)Calculate the work done on the suitcase by the normal force.

d)Calculate the work done on the suitcase by the friction force.

e)Calculate the total work done on the suitcase.

f)If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 3.80 m along the ramp?

Answer 1

1) Expression for work done:

work = F x d = 169 N x 3.80 m = 642.2 J

Round off the result to 3 significant digits

work = 642 J

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2) Work done by gravity:

work = mgh = 18kg * 9.8m/s² * 3.8m * sin34º = 374.8 J

Round off the result to 3 significant digits

work = 375 J

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3) Work done by normal force is always zero.

W = 0 J

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4) Expression for work done by frictional force.

friction work = µmgdcosΘ

work = 0.250 * 18kg * 9.8m/s² * 3.80 m * cos34º = 138.9 J

Round off the result to 3 significant digits

work = 139 J

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5) total work is algebriac sum of all above works. Here, gravity and frcitional forces acting oppoiste to direction of motion. Therefore,

Net work done = (642.2 J - 374.8 - 138.9) J = 128.4 J

Round off the result to 3 significant digits

work = 128 J

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6) Apply work energy theorem, that is work done is equal to net kietic energy.

KE = 128.4 J = ½ mv² = ½ * 18kg * v²

Solve the equation for velocity

v = 3.777 m/s

Round off the result to 3 significant digits

work = 3.78 m/s