A luggage handler pulls a suitcase of mass 16.7 kg up a ramp inclined at an angl

Question

A luggage handler pulls a suitcase of mass 16.7 kg up a ramp inclined at an angle 21.0 above the horizontal by a force of magnitude 155 that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.261. The suitcase travels a distance 4.20 along the ramp. Part A Calculate the work done on the suitcase by the force . Part B Calculate the work done on the suitcase by the gravitational force. Part C Calculate the work done on the suitcase by the normal force. Part D Calculate the work done on the suitcase by the friction force. Part E Calculate the total work done on the suitcase. Part F If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 4.20 along the ramp?

Explanation / Answer

The parallel component of weight: 66 N the perpendicular component of weight: 180 N the kinetic fiction: 49N => 155 > 66 + 49 => it can move up the ramp Work: A = F.s = 155*4.20 = 651 J b) Work = force * displacement Force of friction = mu * normal force = mu mg cos theta Work = mu mg distance cos theta They give you the coefficient (mu), the mass (m), and the angle (theta). You know g. Plugnchug. This should actually be negative because the force opposes the displacement. c) Work of pull = force * displacement This is trivial Total work = work of pull minus the work of friction d) Change in kinetic energy = 1/2 mv^2 = total work so v = sqrt (2 * total work / m)

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