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

The potential energy of a \(1 ~\text{kg}\) particle free to move along the \(x\text-\)axis is given by \(U(x)=\left(\frac {x^4}{ 4}-\frac {x^2}{ 2}\right)~\text J.\) The total mechanical energy of the particle is \(2~\text J.\) Then the maximum speed (in \(\text{ms}^{-1}\)) will be:
1. \(\dfrac{3}{\sqrt{2}} \)
2. \(\sqrt{2}\)
3. \(\dfrac{1}{\sqrt{2}}\)
4.  \(2\)

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. \(\dfrac{Fh}{2}\)
4. \(4Fh\)

1. \(u^{2} \sin^{2}\alpha\)

2. \(\dfrac{m u^{2} \cos^{2} \alpha}{2}\)

3. \(\dfrac{m u^{2}\sin^{2} \alpha}{2}\)

4. \(- \dfrac{m u^{2}\sin^{2} \alpha}{2}\)

A body of mass 'm' is released from the top of a fixed rough inclined plane as shown in the figure. If the frictional force has magnitude F, then the body will reach the bottom with a velocity: $(\mathrm{L}=\sqrt{2}\mathrm{h})$

A body constrained to move along the \({z}\)-axis of a coordinate system is subjected to constant force given by \(\vec{F}=-\hat{i}+2 \hat{j}+3 \hat{k}\) where \(\hat{i},\hat{j} \) and \(\hat{k}\) are unit vectors along the \({x}\)-axis, \({y}\)-axis and \({z}\)-axis of the system respectively. The work done by this force in moving the body a distance of \(4~\text m\) along the \({z}\)-axis will be:
1. \(15~\text J\) 
2. \(14~\text J\) 
3. \(13~\text J\) 
4. \(12~\text J\) 

The potential energy of a particle varies with distance \(r\) as shown in the graph. The force acting on the particle is equal to zero at:

1. \(P\)
2. \(S\)
3. both \(Q\) and \(R\)
4. both \(P\) and \(S\)

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.  $\frac{\mathrm{mgh}}{3}$

A particle is moving such that the potential energy U varies with position in metre as U (x) = ($4{\mathrm{x}}^2$ - 2x + 50) J. The particle will be in equilibrium at: 
1. x = 25 cm
2. x = 2.5 cm
3. x = 25 m
4. x = 2.5 m