Question

A man $(1)$ of mass $m$ stands on an elevator moving with upward acceleration $a .$ A man $(2)$ is standing on the elevator. The elevator starts with the initial velocity $v_{0}$ at time $t = 0 .$ Consider time interval $t$ from the beginning.

(a) What is the work done by normal contact force and gravity on the man $(1)$ as observed by man $(2)$ standing on the elevator and man $(3)$ standing on the ground?

(b) What is the net work done by normal contact force between man $(1)$ and elevator?

Accepted Answer

(a) Observation of man $(1)$ from man $(2)$ . Forces on man are $m g ↓$ and $N ↑$ and pseudo force $m a ↓$ .

Since the displacement of the man $(1)$ relative to man $(2)$ is zero, the forces do not perform work relative to the man $(2)$ . Observation of man $(1)$ from man $(1)$ . Forces on man $(1)$ as seen from are the man $(3)$ are $m g ↓$ and $N ↑$ .

However, the displacement of man $(1)$ relative to man $(3)$ (ground) is not zero. Hence, the forces will perform non-zero work relative to the ground man $(3)$ .

Work done by normal reaction: $W_{N} = \vec{N} \cdot \vec{s} . . . (i)$

Force equation: $N - m g = m a \Rightarrow N = m (g + a) . . . (ii)$

Displacement of man $(1)$ as seen from man $(3)$ ,

$\vec{s} = {\vec{v}}_{13} t + \frac{1}{2} {\vec{a}}_{13} t^{2} = (v_{0} t + \frac{1}{2} a t^{2}) . . . (iii)$

Using the equations $(i), (ii),$ and $(iii),$ we have

$W_{N} = m (g + a) (v_{0} t + \frac{1}{2} a t^{2})$

Similarly, work done by gravity on man $(1)$ as seen from man $(2)$

$W_{g r} = m (\vec{g} \cdot \vec{s}) = (m g) (s) \cos 180 °$

$= - m g (v_{0} t + \frac{1}{2} a t^{2})$

(b) Here net work done means total work done normal reaction on man $(1)$ $(↑)$ and elevator $(↓) .$ As the relative displacement between the points of application of the normal reaction between the person $(1)$ and the elevator is zero, then, as a whole, the sum of work done by the normal reaction is zero.

B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 11: ILLUSTRATION 8.11

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 11: ILLUSTRATION 8.11

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