H.W. Question: If a person with a mass of 54.9 kg rode a distance of 54.6 m on an escalator?

Question

How much work did the escalator do on the individual if it has an inclination of 31.5 degrees? answer in J

acceleration of gravity is 9.8 m/s^2

Answer 1

Work = force * distance (with extra understanding here http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/energy/u5l1aa.html)

So the work done is
F=ma
F=(54.9*9.8)kgm/s^2
= 538.02 N
W=F.d
= 538 * (54.6 sin 31.5)
= 15.35 kJ

The distance that the force is acting on is the vertical, which is why you are getting the vertical component of the escalator movement.

We are assuming no (or negligible) work done horizontally since there is no force given in that direction.

Hope that helps.

Answer 2

Well, the work, in this case, done by the escalator is a function of the force and the displacement. W=Fd.

Because the force in this problem is actually the weight of the person (remember F=ma, where a=g for weight), it is the force of gravity acting on the mass of the person.

And gravity is a conservative force (path-independent)...it doesn't care how you get from point A to point B, it only cares about the net distance, or displacement.

Since you have the angle and the total distance, the vertical displacement would be 54.6m*sin(31.5degrees).

You have F, and you have d. Now solve.

Answer 3

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