Work, Energy, Power and Collisions Questions and Answers

Answer: A

The spring has least potential energy when it is neither compressed nor pulled out and is lying on the ground.

A spring can store two kinds of potential energy: elastic potential energy when it is compressed or stretched from its natural length, and gravitational potential energy when it is raised to a height. A spring at its natural length has no elastic energy, and on the ground it has the least gravitational energy, so this position has the least of both.

A compressed spring on the ground still stores elastic energy, and the springs placed at a height add gravitational energy as well, so options 2, 3 and 4 all have more.

Answer: C

Both Statements 1 and 2 are true.

Statement 1: mechanical energy is kinetic plus potential energy. A body at rest on a shelf has zero velocity, so zero momentum, but it still has potential energy, so its mechanical energy is not zero.

Statement 2: work is done only when the force causes a displacement along its direction. A person pushing a wall exerts a force but does no work, and a force perpendicular to the motion (such as the force keeping a body in a circle) also does no work. Since both statements hold, the options calling either false are wrong.

Answer: C

The kinetic energy is \(\frac{p^{2}}{2m}\).

Express kinetic energy in terms of momentum p = mv:

\[ \begin{aligned} KE &= \tfrac{1}{2}mv^{2} = \frac{m^{2}v^{2}}{2m} \\ &= \frac{p^{2}}{2m} \end{aligned} \]

The other expressions have the wrong units: p²m and 2p²m multiply by mass instead of dividing, and pm is not an energy at all.

Answer: A

Kinetic energy is the energy of motion, and in this classification it includes the radiant energy of electromagnetic waves, thermal energy (the random motion of atoms and molecules), electrical energy (moving charges), sound and mechanical motion. Nuclear energy is the energy stored in the bonds of the nucleus, which is a form of potential energy, along with chemical, gravitational and elastic energy. So nuclear energy is the one that is not kinetic. The answer is option A.

Answer: D

Energy exists in forms such as electrical energy (electricity), heat (thermal energy) and motion (kinetic energy), and others such as light, chemical and nuclear energy. Humidity is the amount of water vapour in the air, which is a property of the atmosphere, and it is not a form of energy. The answer is option D.

Answer: C

The power required is 32 W.

The work done equals the gain in kinetic energy, and the average power is that work divided by the time taken:

\[ \begin{aligned} \Delta K &= \tfrac{1}{2} \times 20 \times (6^{2} - 2^{2}) \\ &= 10 \times 32 = 320\ \text{J} \\ P &= \frac{320}{10} = 32\ \text{W} \end{aligned} \]

Forgetting the half in \(\frac{1}{2}mv^2\) gives 64 W. The work depends on the change in \(v^2\), 36 − 4, not on the change in \(v\), and 54 W and 72 W do not follow from it.

Answer: C

The work done is 9.6 J.

By the work-energy theorem, net work equals the change in kinetic energy. At \(x = 0\), \(v = 2\) m/s; at \(x = 2\) m, \(v = 2(4) + 2 = 10\) m/s.

\[ \begin{aligned} W &= \tfrac{1}{2}m\left(v_{2}^{2} - v_{1}^{2}\right) \\ &= \tfrac{1}{2}(0.2)(100 - 4) = 9.6\ \text{J} \end{aligned} \]

4.8 J is half the correct value (the \(\frac{1}{2}\) applied twice), and 10.8 J or 5.4 J do not come from the change in kinetic energy. Integrating \(F = mv\frac{dv}{dx}\) from 0 to 2 m gives the same 9.6 J.

Answer: B

It is a perfectly inelastic collision.

When the bodies stick together and move with a common velocity, the coefficient of restitution is zero and the loss of kinetic energy is the largest possible consistent with conservation of momentum. That is the definition of a perfectly inelastic collision.

In a perfectly elastic collision kinetic energy is conserved and the bodies separate; partially elastic or inelastic collisions also end with the bodies separating. Energy is transferred in every collision, so the last option is wrong.

Answer: B

The power is proportional to \(t\).

Starting from rest with constant acceleration \(a = F/m\), the velocity is \(v = at\). The power delivered by the constant force is

\[ P = Fv = F \cdot \frac{F}{m}t = \frac{F^{2}}{m}t \]

which grows linearly with time. \(t^{2}\) is how the kinetic energy (the work done) grows, and \(t^{3/2}\) or \(t^{1/2}\) would come from mixing up displacement and velocity.

Answer: E

The INCORRECT statement is that work done depends on the time taken to complete the work.

Work is force times displacement in the direction of the force, W = Fs cosθ. Time does not appear in it; the rate of doing work, power, is what depends on time.

The other statements are correct. In uniform circular motion the centripetal force is perpendicular to the velocity, so it does no work; in dynamic equilibrium the net force is zero, so the net work is zero; gravity does positive work on a falling body; and the work done in stretching a string is positive and stored as elastic energy.