Work, Energy and Power

Work

Work is said to be done by a force when the force produces a displacement on a body on which it acts. \[W=\vec{F}.\vec{S}\] Here, W is the work done by the force \(\vec{F}\) while displacing a body through \(\vec{s}\). Work is a scalar quantity; it has no property of direction but only magnitude.

If the angle between the displacement vector and force vector is \(\theta\) as in fig., then the work done is, \[W=Fscos\theta \hspace{0.1cm}(\because \vec{a}.\vec{b}=abcos\theta)\] Equivalently, W= component of force along the displacement \(\times\) the displacement

Special cases:

When \(\theta=0^\circ\), W=Fs (maximum work done).
When \(\theta=90^\circ\), W=0 (no work done )
When \(\theta < 90^\circ\) ( positive work done )
When \(90^\circ < \theta \leq 180^\circ\) ( negative work done )

\(\implies\) When a coolie travels on a platform with a load on his head, he exerts a vertical force on the load as in fig. . In this case, the angle \(\theta\) between the force vector and the displacement vector is \(90^\circ\) and thus \[W=Fscos\theta=0\] Hence, the work done by the coolie is zero.

\(\implies\)When you slide a body on a rough surface, the direction of the frictional force is opposite to that of the displacement vector \(i.e., \theta=180^\circ\) as in fig. . So, the work done by the frictional force is negative. i.e., \[W= -\hspace{0.1cm} \text{ve}\]

Units of Work : The SI unit of force is Nm or Joule and its CGS unit is erg. \[1 \hspace{0.1cm} \text{J} = 10^7 \hspace{0.1cm}\text{erg}\]

How much work does the force \(\vec{F}=2 \hat i - 6 \hat j\) N do on the block during the displacement \(\vec{s}=-3 \hat i\) m?
\(\begin{align*} W&=\vec{F}.\vec{s}\\ &=(2 \hat i - 6 \hat j).(-3 \hat i)\\ &=-6 \hspace{0.1cm} J \hspace{0.1cm} \because \hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1 \end{align*}\) Thus, the force does the negative work of 6 J on the block.

Work done by a variable force

Consider that the force is applied in the positive direction of x axis and the magnitude of force varies with the position x . fig.(a) shows the plot of such a one-dimensional variable force. We intend to find the work done on the particle by this force as the particle moves from an initial point \(x_i\) to a final point \(x_f\). But, we cannot use \(W=Fscos\theta\) because it applies only for a constant force \(\vec{F}\).
Let's divide the area under the curve of fig.(a) into a number of narrow strips of width \(\Delta x\) which is sufficiently enough to permit us to take the force F(x) reasonably constant.Then, the work done by the force during the displacement \(\Delta x\) is, \[\Delta W = F \Delta x\] To approximate the total work W done by the force as the particle moves from \(x_i\) to \(x_f\), we sum the areas of all the strips between \(x_i\) and \(x_f\) in fig. (b). \[W=\sum \Delta W = \sum F \Delta x\]
Let the strip width approach zero (i.e., \(\Delta x \rightarrow 0\)), then the number of strips become infinitely large and we have, as an exact result, \[W=lim_{\Delta x \rightarrow 0} \sum F \Delta x\] This is equal to the integral of the function F( x ) between the limits \(x_i\) and \(x_f\). Thus, \[W=\int^{x_f}_{x_i} F(x) dx\] Thus, the work done by a variable force is numerically equal to the area between the force curve and the x axis.

A force F = (10+0.50x) acts on a particle in the x direction, where F is in newton and x is in meter. Find the work done by this force during a displacement from x= 0 to x = 2.0 m.
As the force is variable, the work done is then, \[W=\int^{x_f}_{x_i} F(x) dx\] Here, \(x_i=0\) and \(x_f=2\)m, then , \[\begin{align*} W&=\int^{2}_{0} (10+0.50x) dx\\ &=\int^{2}_{0} 10 dx +\int^{2}_{0} 0.50xdx\\ &=\left[10x+0.50\frac{x^2}{2}\right]^2_0\\ &=10(2-0)+0.50 \times \frac{2^2}{2}\\ W&=21 \hspace{0.1cm} \text{J} \end{align*}\]

Energy

Energy is the capacity to do work. It is a scalar quantity and has same unit as that of the work.

Potential energy

It is the energy that can be associated with the configuration (arrangement) or position of a system of objects that exerts force on each other. For example, when a bungee-cord jumper dives from a platform, the system of objects consists of Earth and the jumper. The force between the object is the gravitational force. Here, the configuration of the system changes (the separation between the jumper and the Earth decreases). Here, the energy associated with the state of separation between two objects that attract each other by the gravitational force is thus the gravitational potential energy .
The gravitational potential energy is given as, \(mgh\), for a body of mass m at an height of h .

Kinetic energy

It is the energy associated with the state of motion of an object . The faster the object moves, the greater is its kinetic energy. For a stationary body, kinetic energy is zero.
For an object of mass m with speed v , the kinetic energy is given as, \(\frac{1}{2}mv^2\).

An object of A of mass 10 kg is moving with a velocity of 6 m/s. Calculate its kinetic energy and momentum.
If a constant opposing force of 20 N suddenly acts on A, find the time it takes to come to rest and the distance through which it moves.
Kinetic energy is, \[K.E. =\frac{1}{2}mv^2=\frac{1}{2}\times 10 \times 6^2=180 \hspace{0.1cm} \text{J}\] Momentum is , \[p=mu=10\times 6=60 \hspace{0.1cm} \text{kgm/s}\] Since, the constant opposing force of 20 N acts on a body, the retardation is calculated from \(F=ma\) as, \[a=\frac{F}{m}=\frac{-20}{10}=-2 \hspace{0.1cm}m/s^2\] As the body is brought to rest, it's final velocity v=0 and thus , \[\begin{align*} v&=u+at\\ 0&=6-2t\\ \therefore t&=3 \hspace{0.1cm} s \end{align*}\]

Work - Energy Theorem

Work - Energy theorem states that, " Total work done by a force acting on a body is total change in its kinetic energy."

Consider the body of mass m moving with a initial velocity u on a smooth horizontal surface as in fig. . Let, the force F acts on a body from point A to point B (i.e, the displacement of s) such that its velocity increases to v . The work done by the force is, \[W=Fs\] Since, \(F=ma\), \(W=mas\) and from the equation of motion \(v^2=u^2+2as\), we find, \[v^2-u^2=2as \implies as=\frac{v^2-u^2}{2}\] So, the expression for work done becomes, \[\begin{align*} W&=m\frac{v^2-u^2}{2}\\ &=m\frac{v^2}{2}-m\frac{u^2}{2}\\ &=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \therefore W&=(K.E.)_f-(K.E.)_i \end{align*}\] Here, \((K.E.)_f\) is the final kinetic energy and \((K.E.)_i\) is the initial kinetic energy.Therefore, the work done by the force F is equal to the change in kinetic energy. This proves the work-energy theorem .

You throw a 20 N rock vertically into the air from ground level. You observe that when it is 15 m above the ground, it is travelling at 25 m/s upward. Use the work-energy theorem to find (i) its speed as it left the ground and (ii) its maximum height.
Here, when h=15 m, v = 25 m/s. To find (i) u=? (ii)\(h_{max}\)=?
From work-energy theorem, \[\begin{align*} W&=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ mgh&=\frac{1}{2}m(25)^2-\frac{1}{2}mu^2\\ gh&=\frac{625}{2}-\frac{u^2}{2}\\ -10\times 15&=312.5-\frac{u^2}{2}\\ u^2&=(150+312.5)\times 2\\ \therefore u&=30.41 \hspace{0.1cm} m/s \end{align*}\] For maximum height, final velocity v=0 and thus , \[\begin{align*} W&=0-\frac{1}{2}mu^2 \hspace{0.1cm} \because v=0\\ gh_{max}&=\frac{1}{2}u^2\\ h_{max}&=-\frac{u^2}{2g}\\ \therefore h_{max} &=46.24 \hspace{0.1cm} m \end{align*}\]

Principle of conservation of energy

It states that, energy of an isolated system is constant. Let's understand the principle of conservation of energy by taking an analogous example . Think of many types of energy as being numbers representing money in many types of bank accounts. Rules have been made about what such money numbers mean and how they can be changed. You can transfer money numbers from one acount to another or from one system to another, perhaps electronically with nothing material actually moving. However, the total amount can always be accounted for: It is always conserved (Walker, Haliday & Resnick, Principles of Physics, 10\(^{th}\) edition, Reprint 2016). In the same way, although the energies can be transformed from one to another, the total energy remains conserved. This is the Principle of conservation of energy .

Consider the body of mass m freely falling under the gravity as in fig.

At point A, the body is at rest and thus,
K.E. of the body = 0 P.E. of the body = \(mgh\) Total energy of the body = K.E. + P.E. = \(0+mgh\)=\(mgh\) At point B, the body is in the height ' h- x ' and at a distance of x from point A.
K.E. of the body = \(\frac{1}{2} m v_B^2\) Since, \[\begin{align*} v_B^2&=v_A^2+2gx\\ v_B^2&=2gx \hspace{0.1cm} \because v_A=0\\ \end{align*}\] Thus, K.E. of the body = \(\frac{1}{2}mv_B^2=\frac{1}{2} m\times 2gx=mgx\) P.E. of the body = \(mg(h-x)\) Total energy of the body = \(mgx+mg(h-x)=mgh\) At point C, the body is just about to touch the ground with velocity \(v_C\),
\[\begin{align*} v_C^2&=v_A^2+2gh\\ v_C^2&=2gh \end{align*}\] K.E. of the body = \(\frac{1}{2}mv_C^2=\frac{1}{2}m \times 2gh=mgh\) P.E. of the body = \(mgh = 0 \hspace{0.1cm} \because h=0\) Total energy of the body = \(mgh+0=mgh\)

Thus, the total mechanical energy of the body remains the same at all points during the free fall of the body. As the body falls from a height 'h', initially it's kinetic energy is zero because of its zero initial velocity but the potential energy is maximum (because of maximum height 'h'). However, when the body begins to fall, its kinetic energy increases due to increase in velocity and potential energy reduces. At last, when the body just reaches the ground, its potential energy become zero (ie. mg.0=0) but the kinetic energy reaches to its maximum value. Whatever is the case, the total energy always manage to be conserved. This is illustrated in a graph below.

Conservative forces

A force is said to be conservative if the work done by or against the force in moving a body depends only on the initial and final positions of the body, and not on the nature of the path followed (i.e., path independence) between the initial and the final positions. All the central forces such as electrostatic forces, magnetic force, gravitational forces, etc. are conservative forces.

Path independence of conservative forces
The net work done by a conservative force on a particle moving around any closed path is zero. Consider the ball in figure: it leaves the launch point with speed u and kinetic energy \(\frac{1}{2}mu^2\).

The gravitational force acting on the ball slows it, stops it, and then causes it to fall back down. When the ball returns to the initial point of launch, it again has a speed of u and same kinetic energy. Thus the gravitational force transfers as much as energy from the ball during the ascent as it transfers to the ball during the descent back to the lauch point. The net work is then zero. Thus, the gravitational force is conservative force .

To clear the concept more on conservative force, lets take an example. Suppose that a particle moves from point a to point b as in fig.(a) along either path 1 or path 2. If only a conservative force act on the particle, then the work done on the particle is the same along the two paths. In symbols, it can be written as, \[W_{ab,1}=W_{ab,2}\] and for the conservative force, the net work done during the round trip (fig.(b)) must be zero ( \(W_{ab,1}+W_{ba,2}=0 )\).

Non - conservative forces

A force is said to be non-conservative if the work done by or against the force in moving a body depends on the path followed between the initial and final positions. Kinetic frictional force, viscous force, induction force in a cyclotron, etc. are non-conservative forces.

Power

Power is defined as the rate of doing work . For a constant force F, power is written as, \[P=\vec{F}.\vec{v}=Fvcos\theta\] Here, v is the linear velocity of a body and \(\theta\) is the angle between the force applied and the velocity of the body. Power is a scalar quantity: it is specified only by magnitude.

Across a horizontal floor, a 102 kg block is pulled at a constant speed of 5.5 m/s by an applied force of 125 N directed 38\(^\circ\) above the horizontal. Calculate the rate at which the force does work on the block.
Rate at which force does work on the block is, \[\begin{align*} \text{Power}(P)&=Fvcos\theta\\ &=125\times 5.5\times cos38^0 \hspace{0.1cm}(\because F=ma)\\ \therefore P&=541.75 \hspace{0.1cm} \text{Watt} \end{align*}\]

A train of mass \(2\times 10^5\) kg moves at the constant speed of 72 km/h up a straight incline up against a frictional force of \(1.28\times 10^4\) N. The incline is such that the train rises vertically 1.0 m for every 100 m travelled along the incline. Calculate the necessary power developed by the train.
Here, mass of train, m = \(2\times 10^5\) kg, speed, v = 72 km/h, frictional force, f = \(1.28\times 10^4\) N, \(sin\theta=\frac{1}{100}\), Power, P = ?

Since the train is moving straight incline up, \[\begin{align*} \text{Net force}&=F-(f+mgsin\theta)\\ 0&=F-(f+mgsin\theta) \hspace{0.1cm} \because v=\text{const.} \implies a=0\\ F&=f+mgsin\theta\\ &=1.28\times 10^4+2\times 10^5\times 10\times \frac{1}{100}\\ \therefore F&=3.28\times 10^4 \end{align*}\] So, the power developed by the train is, \[\begin{align*} P&=Fv\\ &=3.28\times 10^4 \times \frac{72\times 1000}{3600}\\ \therefore P&=656000 \hspace{0.1cm} \text{Watt} \end{align*}\]

Collisions

Collision is the mutual interaction between two particles for a short interval of time so that their momentum and kinetic energy may change . Note that , actual physical contact between two bodies is not necessary for collision to occur.

Elastic collision

The collision in which the linear momentum and kinetic energy are conserved is called elastic collision . Collisions between gas molecules, atomic or sub-atomic particles, etc. are the examples of elastic collisions.

Show that in elastic collision, the relative velocity of approach before collision is equal to the relative velocity of separation before collision.
Consider two objects of masses \(m_1\) and \(m_2\) moving with velocities \(u_1\) and \(u_2\) respectively such that \(u_1>u_2\) collides and \(v_1\) and \(v_2\) be their respective velocities after collision as in fig. .

In elastic collision, the linear momentum and kinetic energy are conserved. From the principle of conservation of linear momentum, \[\begin{align*} m_1u_1+m_2u_2&=m_1v_1+m_2v_2\\ m_1(u_1-v_1)&=m_2(v_2-u_2) ... (i) \end{align*}\] From the conservation of kinetic energy, \[\begin{align*} \frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2&=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\\ m_1(u_1^2-v_1^2)&=m_2(v_2^2-u_2^2)\\ m_1(u_1-v_1)(u_1+v_1)&=m_2(v_2-u_2)(v_2+u_2) ... (ii) \end{align*}\] Dividing eqn (ii) by eqn (i), \[\begin{align*} u_1+v_1&=v_2+u_2\\ u_1-u_2&=v_2-v_1 ... (iii) \end{align*}\] From equation (iii), it follows that in elastic collision, the relative velocity of approach \((u_1-u_2)\) before collision is equal to the relative velocity of recession or separation \((v_2-v_1)\) after collision.

Show that in one dimensional elastic collision, the colliding objects with same mass exchange their velocities.
For this question, you also have to solve upto eqn (iii) and ...
From equation (iii),
\[v_2=u_1-u_2+v_1\] Substituting the value of \(v_2\) on eqn (i), \[\begin{align*} m_1u_1+m_2u_2&=m_1v_1+m_2(u_1-u_2+v_1)\\ m_1u_1+m_2u_2&=m_1v_1+m_2u_1-m_2u_2+m_2v_1\\ (m_1-m_2)u_1+m_2u_2&=(m_1+m_2)v_1\\ v_1&=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2} ... (iv) \end{align*}\] Also, from eqn (iii), \[v_1=v_2-u_1+u_2\] Substituting the value of \(v_1\) in equation (i), we get, \[\begin{align*} m_1u_1+m_2u_2&=m_1(v_2-u_1+u_2)+m_2v_2\\ m_1u_1+m_2u_2&=m_1v_2-m_1u_1+m_1u_2+m_2v_2\\ (m_2-m_1) u_2+2m_1u_1&=(m_1+m_2)v_2\\ v_2&=\frac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2} ... (v) \end{align*}\] For same mass \(m_1=m_2=m\), eq (iv) reduces to, \[v_1=\frac{2mu_2}{2m}=u_2\] Similarly, eqn (v) reduces to, \[v_2=\frac{2mu_1}{2m}=u_1\] So,in one dimensional elastic collision, the colliding objects with same mass exchange their velocities (i.e., \(v_1=u_2\) and \(v_2=u_1\)).

Other special cases are:

When the second body is at rest, i.e., \(u_2=0\), from eqn (iv), \[v_1=\frac{(m_1-m_2)u_1}{m_1+m_2}\]
When \(m_1>>m_2\) and \(u_2=0\), then \[v_1=u_1 \hspace{0.1cm} \text{and} \hspace{0.1cm} v_2 \approx 2u_1\]
When \(m_2>>m_1\) and \(u_2=0\), then \[v_1=-u_1 \hspace{0.1cm} \text{and} \hspace{0.1cm} v_2 \approx 0\]

Inelastic collision

The collision in which the linear momentum is conserved but not the kinetic energy is called inelastic collision. Some of the total kinetic energy is converted into non-recoverable heat or sound energy. A bullet remaining embedded in a target such as wood block is an example of inelastic collision.

Show that in inelastic collision, the ratio of kinetic energy before collision is greater than the kinetic energy after collision (i.e., the energy is lost during collision ).
Consider the body of masses \(m_1\) and \(m_2\) with velocity \(u_1\) and \(u_2=0\) respectively. After the collision, they move with the same velocity v (they stick).
In inelastic collision, the linear momentum is conserved. So,

Initial momentum before collision = Final momentum after collision \[\begin{align*} m_1u_1+m_2.0&=m_1v+m_2v\\ m_1u_1&=(m_1+m_2)v\\ v&=\frac{m_1u_1}{m_1+m_2} ... (i) \end{align*}\] Since, the kinetic energy is not conserved in inelastic collision, Kinetic energy before collision, \(K.E.\) =\(\frac{1}{2}m_1u_1^2\) Kinetic energy after collision, \(K.E.'\) = \(\frac{1}{2}(m_1+m_2)v^2\) Thus , the ratio of kinetic energy becomes, \[\begin{align*} \frac{K.E.}{K.E.'}&=\frac{m_1 u_1^2}{(m_1+m_2)v^2}\\ &=\frac{m_1u_1^2}{(m_1+m_2)\left(\frac{m_1v_1}{m_1+m_2}\right)^2}\\ &=\frac{m_1+m_2}{m_1} \end{align*}\] Here, the numerator is greater than the denominator and thus it is obvious that the result is greater than 1. So, \[\begin{align*} \frac{K.E.}{K.E.'}& \gt 1\\ K.E.& \gt K.E.' \end{align*}\] Hence, the kinetic energy before collision \(K.E.\) is greater than the kinetic energy after collision \(K.E.'\) in inelastic collision.

A stationary mass explodes into two parts of mass 4 kg and 40 kg. The initial kinetic energy of larger mass is 10 J. Find the velocity of the smaller mass.
Consider that the mass M at rest explodes into two parts each of mass \(m_1=4\) kg and \(m_2=40\) kg which then moves with velocities \(v_1\) and \(v_2\) respectively as in figure.

Kinetic energy of larger masss, \(K.E._2\) = 10 J
Then, \[\begin{align*} K.E._2 &=\frac{1}{2}m_2v_2^2\\ 10 &=\frac{1}{2}\times 40 v_2^2\\ v_2^2 &=10\times 2 \times \frac{1}{40} \\ v_2 &=\sqrt{\frac{1}{2}}\hspace{0.1cm} \text{m/s} \end{align*}\] From the conservation of linear momentum, \[\begin{align*} m_1u_1+m_2u_2&=m_1v_1+m_2v_2\\ 0&=4v_1+40v_2\\ v_1&=-\frac{40}{4}v_2\\ v_1&=-7.07 \hspace{0.1cm} \text{m/s} \end{align*}\] Thus, the velocity of the smaller mass is 7.07 m/s. Negative sign indicates that the direction of velocities of two masses are opposite.

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Physics in Depth