A 54 kg woman steps onto an up-going escalator, which has an incline of 32 degre

Question

A 54 kg woman steps onto an up-going escalator, which has an incline of 32 degrees with respect to the horizontal and is moving at 0.5 m/s. The top of the escalator is 25 m above the ground level. (A) Calculate how much work is done by the friction force between the woman's feet and the escalator step, as she moves from the bottom to the top of the escalator. (B) Calculate how much work is done by gravity as she moves from the bottom to the top of the escalator. (C) Calculate how much work is done by the normal force on the woman's feet, as she moves from the bottom to the top of the escalator. (D) What is the total work done on the woman as she moves from the bottom to the top?

Explanation / Answer

A)

Wf = work done by friction = 0

Since the friction force is horizontal and displacement is vertical . hence angle between friction force and displacement is 90 . and since the angle is 90 , work done is 0 since Cos90 = 0

B)

Wg = work done by gravity = - mg h = - 54 x 9.8 x 25 = - 13230 J

c)

Wn = work done by normal force= Fn h = mg h = 54 x 9.8 x 25 = - 13230 J    since Fn = mg

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