The potential energy U of a system is given by $\mathrm{U}= \mathrm{A} - {\mathrm{Bx}}^2$ (where x is the position of its particle and A, B are constants). The magnitude of the force acting on the particle is:

1. Constant

2. Proportional to x

3. Proportional to ${\mathrm{x}}^2$

4. Proportional to $\left(\frac{1}{\mathrm{x}}\right)$

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 particle is moving such that the potential energy U varies with position in metres as U = ($4{\mathrm{x}}^2$ - 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

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:

1. B/A

2. B/2A

3. 2A/B

4. A/B

Potential energy (U) related to coordinates is given by U = 3(x + y). Work done by the conservative force when the particle is going from (0, 0), (2, 3) is:
1.  15 J
2.  -15 J
3.  12 J
4.  10 J

The potential energy of a particle of mass m varies as the magnitude of the $U=a{x}^2+by.$ The magnitude of the acceleration of the particle at (0, 3) is: (symbols have their usual meaning)

1.  $\sqrt{\frac{\mathrm{b}}{\mathrm{m}}}$

2.  $\sqrt{\frac{3\mathrm{b}}{\mathrm{m}}}$

3.  $\frac{\mathrm{b}}{\mathrm{m}}$

4.  Zero

The figure shows the potential energy function U(x) for a system in which a particle is in a one-dimensional motion. What is the direction of the force when the particle is in region AB? (symbols have their usual meanings)

1.  The positive direction of x

2.  The negative direction of X

3.  Force is zero, so direction not defined

4.  The negative direction of y

The potential energy of a particle of mass 1 kg free to move along the X-axis is given by $\mathrm{U}\left(\mathrm{x}\right) = \left(3{\mathrm{x}}^2 - 4\mathrm{x} +\hspace{0.17em}6\right) \mathrm{J}$. The force acting on the particle at x = 0 will be:

1.  2$\hat{\mathrm{i}}$ N

2.  -4$\hat{\mathrm{i}}$ N

3.  5$\hat{\mathrm{i}}$ N

4.  4$\hat{\mathrm{i}}$ N

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 given diagram represents the potential energy curve of a particle in a field. The particle will be in equilibrium at which position(s):

1. B and D

2. A and C

3. A, B, and C

4. A, B, C, and D