Question

A 58 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 23 m above the ground level.

A) Calculate how much work is done by gravity, as she moves from the bottom to the top of the escalator.

Express your answer in Joules to two significant figures.

B)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.

C)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.

Answer 1

Given is :

Weight = m = 58 kg

Angle of escalator = A = 32 degrees

speed of escalator = u = 0.5 m/s

F1 = mg cos(32)

F1 = 58*(9.8)*(0.848) = 482.01 N

F2 = mg sin(32) = 58*(9.8)*(0.5299) = 301.195 N

a) work done by gravity = F2*D

w1 = 301.195*23 =6927.485 J

b) work done by the normal force = F1*D

w2 = 482.01*23 = 11086.23 J

c)

kinetic energy = 1/2(mv²)

ke = (0.5)*58*(0.5*0.5) = 7.25 J

the total work done on the woman = w1 + w2 + ke = 6927.485 + 11086.23 + 7.25 = 18020.965 J