Work, Energy and Power

- (a) When lifting the bucket, the man’s force and the displacement are in the same direction, so work done is positive.
- (b) Gravitational force acts downward while the bucket moves upward, so work done by gravity is negative.
- (c) Friction opposes the motion down the incline, so work done by friction is negative.
- (d) For uniform velocity on a rough horizontal plane, the applied force balances friction and acts along the displacement, so its work is positive.
- (e) Air resistance opposes the motion of the pendulum, so work done by the resistive force is negative.

Given: $m=2\,\text{kg}$, $F=7\,\text{N}$, $\mu_k=0.1$, $g=10\,\text{m s}^{-2}$, $t=10\,\text{s}$.

Friction force:
$$f=\mu_k mg=0.1\times 2\times 10=2\,\text{N}$$
Net force:
$$F_{net}=7-2=5\,\text{N}$$
Acceleration:
$$a=\frac{F_{net}}{m}=\frac{5}{2}=2.5\,\text{m s}^{-2}$$
Distance moved in $10$ s from rest:
$$s=\frac12 at^2=\frac12\times 2.5\times 10^2=125\,\text{m}$$

(a) Work done by applied force:
$$W_F=Fs=7\times 125=875\,\text{J}$$
(b) Work done by friction:
$$W_f=-fs=-2\times 125=-250\,\text{J}$$
(c) Work done by net force:
$$W_{net}=F_{net}s=5\times 125=625\,\text{J}$$
(d) Change in kinetic energy:
$$\Delta K=W_{net}=625\,\text{J}$$
The textbook question expects the work-energy theorem: the net work equals the change in kinetic energy.

Note: The values computed from the question as printed are the correct school-physics result; if your edition’s numbers differ in the margin/printing, the method is the same.

From the graph idea of potential energy, the particle cannot be found where its total energy is less than the potential energy, because then kinetic energy would be negative, which is impossible.

So for each curve in Fig. 5.11:
- Find the regions where **$E < V(x)$; those are forbidden.
- The minimum total energy must be at least equal to the maximum value of $V(x)$** in the allowed region.

The answer depends on the specific shapes in Fig. 5.11, but the rule is:
$$K = E - V(x) \ge 0 \quad \Rightarrow \quad E \ge V(x)$$
Hence the particle can move only where the total energy line lies on or above the potential-energy curve.

For a spring,
$$V(x)=\frac12kx^2$$
Given $k=0.5\,\text{N m}^{-1}$, so
$$V(x)=\frac12(0.5)x^2=0.25x^2$$
At the turning point, all energy is potential, so
$$V(x)=E=1\,\text{J}$$
Thus
$$0.25x^2=1$$
$$x^2=4$$
$$x=\pm 2\,\text{m}$$
So the particle turns back at those positions.

The subparts shown in the chapter are the standard conceptual questions from 5.5 Answer the following. Their answers are:

(a) The heat energy comes at the expense of the rocket’s kinetic energy and internal energy of the rocket, not the atmosphere.

(b) Over a complete orbit, the gravitational force is conservative and depends only on initial and final positions. Since the comet returns to its starting point, the net work is zero.

(c) As the satellite comes closer to Earth, its gravitational potential energy decreases and this lost potential energy is converted into kinetic energy, so its speed increases.

(d) The work done is greater in the case where the man pulls the rope behind him and lifts the hanging mass, because he does work against gravity on the mass. Carrying the mass on his hands does not involve the same increase in the mass’s height during horizontal walking.

The heat for burning is obtained at the expense of the rocket (its energy), not the atmosphere. The textbook’s concept is that friction converts the rocket’s energy into heat.

The gravitational force is conservative. For a comet completing a full elliptical orbit, its initial and final positions are the same, so the net work done by gravity over one complete revolution is zero, even though the force is not always perpendicular to the velocity.

This is because the work done by a conservative force depends only on the end points of the path, and for a closed path the end points coincide.

As the satellite comes closer to Earth, its gravitational potential energy decreases. Due to dissipation, some mechanical energy is lost, but the remaining energy is such that the orbit lies at a smaller radius where the satellite must move faster. In an orbit, a lower orbit corresponds to a larger orbital speed.

So its speed increases progressively as it spirals inward.

In the first case, the man carries the mass horizontally, so its displacement is horizontal while the supporting force is vertical; hence the work done on the mass by the man is zero.

In the second case, the rope goes over a pulley and the hanging mass is raised as the man walks, so the man does positive work against gravity. Therefore, the work done is greater in the second case.

The correct alternatives are:
- (a) decreases
- (b) kinetic
- (c) external force
- (d) total linear momentum

Reasoning:
- A positive conservative force does positive work, so potential energy decreases.
- Work done against friction dissipates kinetic energy.
- The rate of change of total momentum of a system depends on the external force.
- In an inelastic collision, total linear momentum is conserved; kinetic energy is not.

For a conservative force, positive work done by the force means the system loses potential energy:
$$\Delta V = -W$$
So if $W>0$, then $\Delta V<0$. Hence potential energy decreases.

Work done against friction appears as a loss of kinetic energy. Friction opposes motion and reduces the body’s motion energy.

For a many-particle system, the rate of change of total momentum is equal to the external force on the system. Internal forces cancel in pairs by Newton’s third law.

In an inelastic collision, total linear momentum is conserved. Total kinetic energy is not conserved, though total energy of the system remains conserved when all forms are included.

The statements are:
- (a) False
- (b) False
- (c) False
- (d) True

Reasons:
- In an elastic collision, the total momentum and kinetic energy of the system are conserved, but not necessarily each body’s individual momentum and energy.
- Total energy is conserved only when all forms of energy are accounted for; with external work or dissipation, the mechanical energy of the system may change.
- Work over a closed loop is zero only for conservative forces, not every force.
- In an inelastic collision, some kinetic energy is lost to heat, sound, deformation, etc., so final kinetic energy is less than the initial kinetic energy.

In an elastic collision, the total momentum and total kinetic energy of the system are conserved, but the momentum and energy of each body separately need not be conserved. They are exchanged between the two bodies during collision.

Total energy is conserved only in an isolated system when all forms are included. If external forces do work, or if energy leaves/enters the system, the energy of the body/system as defined in the problem need not remain unchanged. So the statement is false.

Work done over a closed loop is zero only for conservative forces. For non-conservative forces like friction, it is not zero. Therefore the statement is false.

In an inelastic collision, part of the initial kinetic energy is converted into other forms such as heat, sound, and deformation. So the final kinetic energy is always less than the initial kinetic energy of the system.

The subparts ask about conservation during collisions:
- kinetic energy during the collision,
- linear momentum during the collision,
- same questions for an inelastic collision,
- and whether dependence of potential energy on separation implies elastic or inelastic.

These are answered in the next entries.