A man(1) of mass `m` strands on an elevator moving with upward acceleration `a`. A man(2) is standing on the elevator. Elevator starts with initial velocity `v_0` at time `t=0`. Consider time interval t from beginning.
a. What is the work done by normal contact force and gravity on the man(1) as observe by man(2) standing on the elevator and man(3) standing on ground?
b. What is the net work done by normal contact force between man(1) and elevator?

Explanation

a. Observation of man(1) from man(2)
Forces on man are `mgdarr` and `N uarr` and pseudo force `ma darr`
Since the displacement of the man(1) relative to man(2) is zero, the force do not perform work relative to the man(2).

Observation of man(1) from man(3)
Forces on man(1) as seen from are man(3) are `mg darr` and `N uarr`

However, the displacement of man(1) relative to man(3) (ground) is not zero.
Hence, the force will perform non-zero work relative to ground (man3).
Work done by normal reaction:
`W_N=vecN.vecs` (i)
Force equation: `N-mg=maimpliesN=m(g+a)`
Displacement of man(1) as seen from man(3),
`vecs=vecvt+1/2veca_13t^2=(v_0t+1/2at^2)`
Using Eqs. (i), (ii), and (iii), we have
`W_N=m(g+a)(v_0t+1/2at^2)`
Similarly, work done by gravity on man(1) as seen from man(2)
`W_(gr)=mvecg.vecs=(mg)(s)cos180^@`
`=-mg(v_0t+1/2at^2)`
b. Here, net work done means total work done normal reaction on man(1) `(uarr)` and elevator `(darr)`. As the relative displacement between the points of application of the normal reaction between person(1) and elevator is zero, then, as a whole, the sum of work done by normal reaction is zero.