A 68 kg woman steps onto an up-going escalator, which has an incline of 32 with

Question

A 68 kg woman steps onto an up-going escalator, which has an incline of 32 with respect to the horizontal and is moving at 0.5 m/s. The top of the escalator is 15 m above the ground level.

A) 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.

Express your answer in Joules to two significant figures.

B) What is the total work done on the woman as she moves from the bottom to the top?

Express your answer in Joules to two significant figures.

Explanation / Answer

A)

work done = change in gravitational potential energy = mg (hf - hi) = (68) (9.8) (15) = 9996 J = 1.0 x 104 J

B)

Total work done = change in gravitational potential energy = mg (hf - hi) = (68) (9.8) (15) = 9996 J = 1.0 x 104 J

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