A body constrained to move along the \(\mathrm{z}\)-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 \(\mathrm{x}\)-axis, \(\mathrm{y}\)-axis and \(\mathrm{z}\)-axis of the system respectively. The work done by this force in moving the body a distance of \(4\) m along the \(\mathrm{z}\)-axis will be:
1. \(15\) J
2. \(14\) J
3. \(13\) J
4. \(12\) J

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: $(\mathrm{L}=\sqrt{2}\mathrm{h})$

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. $\frac{Fh}{2}$

4. 4Fh

The potential energy of a 1 kg particle free to move along the x-axis is given by:

$\mathrm{U}\left(\mathrm{x}\right) = \left(\frac{{\mathrm{x}}^4}{4}-\frac{{\mathrm{x}}^2}{2}\right)\mathrm{J}$

The total mechanical energy of the particle is 2J. Then, the maximum speed (in ms^-1) will be
1. \(3 \over \sqrt{2} \)
2. \(\sqrt{2}\)
3. \(1 \over \sqrt{2}\)
4. 2

A particle moves with a velocity of $(5\hat{\mathrm{i}}-3\hat{\mathrm{j}}+6\hat{\mathrm{k}})$m/s under the influence of a constant force $\overrightarrow{F}=(10\hat{\mathrm{i}}+10\hat{\mathrm{j}}+20\hat{\mathrm{k}})$ N. The instantaneous power applied to the particle is:

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 = 4s is:
      

A particle of mass 'm' is projected at an angle '$\mathrm{\alpha \text{'}}$ with the horizontal, with an initial velocity 'u'. The work done by gravity during the time it reaches its highest point is:

1. $u^2{\sin }^{2}a$

2. $\frac{m{u}^2{\cos }^2\alpha }{2}$

3. $\frac{m{u}^2{\sin }^2\alpha }{2}$

4. $-\frac{m{u}^2{\sin }^2\alpha }{2}$

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:

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