A position dependent force \(F=7-2x+3x^2\) N acts on a small body of mass \(2\) kg and displaces it from \(x = 0\) to \(x = 5\) m. The work done in joule is:

1.	\(70\)	2.	\(270\)
3.	\(35\)	4.	\(135\)

A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on the table is:

1. \(MgL\)

2. \(MgL/3\)

3. \(MgL/9\)

4. \(MgL/18\)

A particle of mass 'm' is moving in a horizontal circle of radius 'r' under a centripetal force equal to –K/r², where K is a constant. The total energy of the particle will be:

1. $\frac{K}{2r}$

2. $-\frac{K}{2r}$

3. $-\frac{K}{r}$

4. $\frac{K}{r}$

The relationship between force and position is shown in the given figure (in a one-dimensional case). The work done by the force in displacing a body from \(x = 1\) cm to \(x = 5\) cm is:

1.   \(20\) ergs
2.   \(60\) ergs
3.   \(70\) ergs
4.   \(700\) ergs

A stone is projected from a horizontal plane. It attains maximum height, 'H', and strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assuming the collision to be elastic, the height of the point on the wall where the ball will strike will be:

A ball is thrown vertically downwards from a height of 20 m with an initial velocity v_o. It collides with the ground, loses 50% of its energy in a collision and rebounds to the same height. The initial velocity v_o is: (Take g = 10 ms^-2)

1. 14 ms^-1
2. 20 ms^-1
3. 28 ms^-1
4. 10 ms^-1

A block of mass M is attached to the lower end of a vertical spring. The spring is hung from the ceiling and has a force constant value of k. The mass is released from rest with the spring initially unstretched. The maximum extension produced along the length of the spring will be:

On a frictionless surface, a block of mass \(M\) moving at speed \(v\) collides elastically with another block of the same mass \(M\) which is initially at rest. After the collision, the first block moves at an angle \(\theta\) to its initial direction and has a speed \(\frac{v}{3}\). The second block’s speed after the collision will be:
1. \(\frac{2\sqrt{2}}{3}v\)
2. \(\frac{3}{4}v\)
3. \(\frac{3}{\sqrt{2}}v\)
4. \(\frac{\sqrt{3}}{2}v\)

The potential energy of a particle in a force field is \(U=\)$\frac{A}{r^2}-\frac{B}{r}$ where \(A\) and \(B\) are positive constants and \(r\) is the distance of the particle from the center of the field. For stable equilibrium, the distance of the particle is:

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: