Chapter 5: Work, Energy, and Power

Ans: Positive.
Work done is the dot product of force and displacement. This means that the work will be positive if the projection of one of them on the other is in the same direction as the other. In other words, if the two vectors make an acute angle, then the work done is positive. Here, the bucket is being lifted upwards, which is in the same direction of the pull-force being applied. Hence the work done is positive.

1. work done by gravitational force in the above case,

Ans: Negative.
The gravitational force is acting downward, and the displacement is upwards, which means the angle these make \[\sim 180{}^\circ \](obtuse). This means that the dot product of the two vectors would be negative.

1. work done by friction on a body sliding down an inclined plane,

Ans: Negative.

The direction of frictional force is opposite to the direction of motion; hence the work done by the frictional force is negative in this case.

1. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,

Ans: Positive.
The frictional force is acting in the direction opposite to the motion of the body. However, to maintain the constant velocity, extra force is applied in the direction of the motion. So this means that the work done by this extra force is positive.

1. work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 

Ans: Negative.

The resistive force always acts in the direction opposite to the motion of the body (pendulum). Thus the work done by this force is also negative.

Ans: Given the mass of the body, \[m=2kg\]

The force applied on the body, \[F=7N\]

The given coefficient of kinetic friction, \[\mu =0.1\]

The initial velocity of the body, \[u=0\]

Time, \[t=10s\]

The acceleration produced in the body by the applied force is given by Newton’s second law

of motion as:

\[a'={}^{F}/{}_{m}\text{=}{}^{7}/{}_{2}\text{=}3.5\text{ }m{{s}^{-2}}\] 

And the frictional force is given as: \[f=\mu mg=0.1\times 2\times 9.8=1.96N\]

The acceleration produced by the frictional force:

\[a''=1.96/2=0.98\text{ }m{{s}^{-2}}\]

Total acceleration of the body is the sum of the two accelerations: \[a'+a''=3.5+0.98\text{ }m{{s}^{-2}}\]

Using the total acceleration, the distance traveled by the body is given by the equation of motion:

\[s=ut+{}^{1}/{}_{2}a{{t}^{2}}=0+{}^{1}/{}_{2}\times 2.52\times {{10}^{2}}=126m\]

Work done by the applied force (F),

\[{{W}_{a}}=F.s=7\times 126=882J\].

Hence the work done by the applied force is 882J.

1. work done by friction in \[10s\], 

Ans: The work done by the frictional force (f), 

\[{{W}_{f}}=f.s=-1.96\times 126=-247J\].

1. work done by the net force on the body in \[10s\], 

Ans: The net force acting on the body\[=7+\left( 1.96 \right)=5.04N\]

Hence the work done by the net force in the 10s, is equal to the product of the force and the distance travelled by the body, \[{{W}_{Net}}=5.04\times 126=635J\].

1. change in kinetic energy of the body in \[10s\], and interpret your results

Ans: Using the first equation of motion, final velocity can be calculated as:

\[v=u+at=0+2.52\times 10=25.2\text{ }m/s\]

The change in kinetic energy is

\[={}^{1}/{}_{2}m{{v}^{2}}-{}^{1}/{}_{2}m{{u}^{2}}={}^{1}/{}_{2}m({{v}^{2}}-{{u}^{2}})=635J\]

Hence the change in the kinetic energy is obtained as 635J.

Solution coming soon.

Ans: As given, the total energy of the particle, \[E=\text{ }1\text{ }J\] 

The potential energy of the particle \[V(x)=k{}^{{{x}^{2}}}/{}_{2}\]

The force constant of the oscillator is \[k=0.5N{{m}^{-1}}\]

The total energy is: \[E=V+T\] where T is the Kinetic energy of the particle. According to the conservation law, this energy does not change throughout the motion.

\[\Rightarrow 1={}^{1}/{}_{2}m{{v}^{2}}+{}^{1}/{}_{2}k{{x}^{2}}\]

When the pendulum ‘turns back’, velocity (and hence T) becomes zero. 

\[\Rightarrow 1={}^{1}/{}_{2}k{{x}^{2}}\]

\[\Rightarrow x=\pm 2\]

Hence, the particle turns back when it reaches \[x=\pm 2\].

Ans: Rocket.

Total energy is given by the relation:

$TE=PE+KE=\frac{1}{2}m{{v}^{2}}+mgh$

Now, when the casing of a rocket burns due to friction, the mass of the rocket gets reduced, which further causes the total energy to reduce. Clearly, it is at the cost of the rocket that the heat needed for burning is generated.

1. Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

Ans: Gravitational force is a conserved force, which means that the work done by this force is independent of the path taken by the object. This means that if an object comes back to a place where it started from, i.e. when the total displacement is zero (i.e., a closed path), the work done is zero.

1. An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

Ans: The atmospheric resistance costs the satellite its total energy. This means that the total energy keeps decreasing. So even when the kinetic energy increases due to the increase in the speed, the potential energy decreases much faster, leading to an overall drop in the energy. 

So basically, the energy of the satellite is given as the sum of its kinetic energy, its potential energy, and the energy it loses to the atmospheric resistance.

1. In (i), the man walks \[\mathbf{2m}\] carrying a mass of \[15kg\] on his hands. In (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley and a mass of \[15kg\] hangs at its other end. In which case is the work done greater? 

Ans: In the case where the person is carrying the weight horizontally, which means he has applied the force (F) in the vertically upward direction, but the displacement (s) is horizontal. The work done (W) is given by \[W=F.s=Fs\cos \theta \] and the \[\theta \] is the angle between the two vectors. Since \[\theta =90{}^\circ \]the work becomes zero. 

Similarly, in the second case, the person is pulling the rope horizontally and the motion of the weight is upwards, and hence the two vectors are perpendicular to each other. Hence the work is zero.

Ans: decreases.

The body moves towards the lower potential hence decreasing the potential energy of the body. 

1.  Work done by a body against friction always results in a loss of its kinetic/potential energy.

Ans: Kinetic energy.

Friction usually reduces the velocity of the body and hence reduces the Kinetic energy as well.

1.  The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.

Ans: External force.

Internal forces cannot produce any change in the total momentum of a body. It’s a fact that the sum of all the internal forces is zero. Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system.

1. In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies

Ans: Total linear momentum.

The momentum can neither be created nor destroyed. However the energies can change their forms.

Ans: False

Law of conservation of momentum states that the total momentum of the system is conserved when there is no net external force applied on the system. The momentum of one body can be transferred to other body in the system.

1.  Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

Ans: False

Any external force on a body can change the total energy of the body.

1.  Work done in the motion of a body over a closed loop is zero for every force in nature.

Ans: False

Total energy over a closed-loop motion is only zero if the force involved is conservative. Eg. The friction would do zero work in a closed-loop.

1.  In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Ans: True

In inelastic collisions, the total energy is usually lost as sound or heat or vibrations which means that the initial kinetic energy is more than the final kinetic energy

Ans: No

At the point of contact, the moving billiard ball drops its velocity to zero and is still in the process of transferring its momentum and energy to the next ball. At this moment, the ball, which came to a sudden stop after hitting the next ball, has transferred its kinetic energy to the potential energy (of vibration), which is then transferred to the next ball as its kinetic energy.

1. Is the total linear momentum conserved during the short time of an elastic collision of two balls?

Ans: Yes

In an elastic collision, the total linear momentum is always conserved at every point in time.

1. What are the answers to (a) and (b) for an inelastic collision?

Ans: No, Yes.

There is a definite loss of kinetic energy in inelastic collisions, which happens during the collision. This is another reason why the total kinetic energy is not conserved.

The law of conservation of momentum is applicable to both elastic and inelastic collisions. Hence the total linear momentum is always conserved throughout the collision.

1. If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Ans: Elastic.

If the potential energy is only dependent on the separation distance, this means that the induced force is independent of the path that is taken by the balls. Hence the force is conservative. This means that the collisions here are also going to be elastic.

Ans: It is given that the body is accelerated with a constant value-\[a\] hence the corresponding force experienced is \[F=ma\].

Let’s say that the initial velocity of the body is \[u\].

Thus, the velocity- \[v\] of the body at any given point in time- \[t\] can be written as \[v-u=a{{t}^{{}}}\]

And the power is

\[P=F.v=(ma)v\]

\[\Rightarrow P=ma(at+u)=m{{a}^{2}}t+mau\]

The mass, acceleration, and initial velocity are constant.

\[\Rightarrow P\propto t\]

Hence it can be noted that power is linearly dependent on t.

Ans: In this case, the power is constant:

\[P=F.v=(ma)v\]

Since the mass of the body doesn’t change, the acceleration and the velocity are inversely proportional.

\[\Rightarrow a=\frac{F}{mv}\]

Also,

\[\frac{dv}{dt}=a\]

Equating the two accelerations

\[\frac{dv}{dt}=a=\frac{F}{mv}\]

\[\Rightarrow \frac{m}{F}\int{vdv}=\int{dt}\]

\[\Rightarrow \frac{m}{F}{{v}^{2}}=t+C\]

\[\Rightarrow v\propto {{t}^{1/2}}\]

Hence the option (i) is valid.