The minimum work done in pulling up a block of wood weighing \(2\) kN for a length of \(10\) m on a smooth plane inclined at an angle of \(15^\circ\) with the horizontal is (given: \(\mathrm{sin}15^\circ=0.2588)\):
1. \(4.36\) kJ
2. \(5.17\) kJ
3. \(8.91\) kJ
4. \(9.82\) kJ

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 of a particle \((x)\) varies with time \((t)\) as \(x = (t - 2)^2\), where \(x\) is in meters and \(t\) is in seconds. Calculate the work done during \(t=0\) to \(t=4\) s if the mass of the particle is \(100~\text{g}.\)
1. \(0.4~\text{J}\)
2. \(0.2~\text{J}\)
3. \(0.8~\text{J}\)
4. zero

The position-time \((x\text- t)\) 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

A bicyclist comes to a skidding stop in \(10\) m. During this process, the force on the bicycle due to the road is \(200\) N is directly opposed to the motion. The work done by the cycle on the road is:

A cord is used to vertically lower a block of mass m by a distance d at a constant downward acceleration of $\frac{\text{'}\mathrm{g}\text{'}}{4}$. The work done by the chord on the block will be:

1.  $\frac{3}{4}$mgd

2.  -$\frac{3}{4}$mgd

3.  $\frac{1}{4}$mgd

4.   -$\frac{1}{4}$mgd