The potential energy \(\mathrm{U}\) of a system is given by (where \(\mathrm{x}\) is the position of its particle and \(\mathrm{A},\) \(\mathrm{B}\) are constants). The magnitude of the force acting on the particle is:
1. constant
2. proportional to \(\mathrm{x}\)
3. proportional to
4. proportional to
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 metre as U (x) = ( - 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
1. | \(\frac{B}{A}\) | 2. | \(\frac{B}{2A}\) |
3. | \(\frac{2A}{B}\) | 4. | \(\frac{A}{B}\) |
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 \(U(x) = (3x^2-4x+6)~\text{J}\). The force acting on the particle at x = 0 will be:
1. 2 N
2. -4 N
3. 5 N
4. 4 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.
2.
3.
4.