A 63.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of 4.20 m. What is the work done on the man by (a) the gravitational force and (b) the escalator

Explanation:

Given that,

Mass of man =Mm=63kg

W=mg

W=63×9.81

W=618.03N

g=9.81m/s²

The escalator moves at a constant velocity, then this shows that it is not accelerating, then, a=0m/s²

Height escalator traveled

H=4.2m.

We assume that the weight of the escalator is negligible.

a. Work done by gravity

Work done by gravity is given as

Work done(gravity)=mgh

Since the work is done against gravity, then, g is negative

Then,

Work done(gravity)=-mgh

Work done(gravity)=-63×9.81×4.2

Work done(gravity)=-2595.726J

b. Work done by escalator

Using equation of motion to know the force pulling the escalator upward

ΣF = ma, but a=0

ΣF = 0

Only two force is acting on the in y axis, the Normal force and the weight

N-W=0

N=W

Since W=618.03N

Then, N=618.03

The normal is the force pulling the escalator upward

Then, the work done by escalator is given as

Work done=Force × distance

Work done=618.03×4.2

Work done= 2595.76J

Work done by escalator is 2595.76J

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

it has to be b, about 97 percent.