A force of 5 N making an angle $\theta$ with the horizontal acting on an object displaces it by 0.4 m along the horizontal direction. If the object gains kinetic energy of 1 J then the component of the force is:

The bob of a simple pendulum having length l, is displaced from the mean position to an angular position θ with respect to vertical. If it is released, then the velocity of the bob at the lowest position will be:

1. $\sqrt{2gl(1-\cos \theta )}$

2. $\sqrt{2gl(1+\cos }$ $\theta )$

3. $\sqrt{2gl}$ $\cos \theta$

4. $\sqrt{2gl}$

A block is carried slowly up an inclined plane. If ${\mathrm{W}}_{\mathrm{f}}$ is work done by the friction, ${\mathrm{W}}_{\mathrm{N}}$ is work done by the reaction force, ${\mathrm{W}}_{\mathrm{g}}$ is work done by the gravitational force and ${\mathrm{W}}_{\mathrm{ex}}$ is the work done by an external force, then choose the correct relation(s):

1.  ${\mathrm{W}}_{\mathrm{N}}$ $+$ ${\mathrm{W}}_{\mathrm{f}}$ $+$ ${\mathrm{W}}_{\mathrm{g}}$ $+$ ${\mathrm{W}}_{\mathrm{ex}}$ $=$ $0$

2.  ${\mathrm{W}}_{\mathrm{N}}$ = 0

3.  ${\mathrm{W}}_{\mathrm{ex}}$ $+$ $$ ${\mathrm{W}}_{\mathrm{f}}$ $=$ $-{\mathrm{W}}_{\mathrm{g}}$

4.  All of these

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 block is released from rest from a height of h = 5 m. After travelling through the smooth curved surface, it moves on the rough horizontal surface through a length l = 8 m and climbs onto the other smooth curved surface at a height h'. If $\mathrm{\mu }$ = 0.5, find h'.

A block of mass \(m\) is connected to a spring of force constant \(K\). Initially, the block is at rest and the spring is relaxed. A constant force \(F\) is applied horizontally towards the right. The maximum speed of the block will be:

                              
1. \(\frac{F}{\sqrt{2mK}}\)
2. \(\frac{\sqrt{2}F}{\sqrt{mK}}\)
3. \(\frac{F}{\sqrt{mK}}\)
4. \(\frac{2F}{\sqrt{2mK}}\)