A person-1 stands on an elevator moving with an initial velocity of 'v' & upward acceleration 'a'. Another person-2 of the same mass m as person-1 is standing on the same elevator. The work done by the lift on the person-1 as observed by person-2 in time 't' is:

1.  $\mathrm{m}\left(\mathrm{g}\right)$ $+$ $\mathrm{a}\left(\mathrm{vt}\right)$ $+$ $\frac{1}{2}{\mathrm{at}}^2$

2.  $-\mathrm{mg}\left(\mathrm{vt}\right)$ $+$ $\frac{1}{2}{\mathrm{at}}^2$

3.  0

4.  $\mathrm{ma}\left(\mathrm{vt}\right)$ $+$ $\frac{1}{2}{\mathrm{at}}^2$

A block of mass m is placed in an elevator moving down with an acceleration $\frac{\mathrm{g}}{3}$. The work done by the normal reaction on the block as the elevator moves down through a height h is:

1.  $\frac{-2\mathrm{mgh}}{3}$

2.  $\frac{-\mathrm{mgh}}{3}$

3.  $\frac{2\mathrm{mgh}}{3}$

4.  $\frac{\mathrm{mgh}}{3}$

In the diagram shown, force \(F\) acts on the free end of the string. If the weight \(W\) moves up slowly by distance \(h,\) then work done on the weight by the string holding it will be: (pulley and string are ideal)

1. \(Fh\)
2. \(2Fh\)
3. \(Fh/2\)
4. \(4Fh\)

The position-time graph of a particle of mass \(2\) kg is shown in the figure. Total work done on the particle from \(t=0\) to \(t=4\) s is:
                   
1. \(8\) J
2. \(4\) J
3. \(0\) J
4. can't be determined