Work Done in Physics

What is Work Done in Physics?

Work done in Physics is entirely different from our usual understanding of day-to-day work like working in an office, doing workouts, etc.

Work done in physics should be taken more as a scientific/physical definition rather than correlating it with real life. Although many similarities exist, strictly speaking, these are different.

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To better understand work done in Physics, let us see its formula.

Work done equation

The work done by a constant force F when its point of application undergoes a displacement S is defined to be

$W = FS\cos \theta$

Here, $\theta$ is the angle between F and S.

Only the component of F along S that is $F\cos \theta$ contributes to the work done. Strictly speaking, the work is done by the source or agent that applies the force.

To understand it better, let us calculate the work done in the below examples.

Example 1

If we lift a body from rest to a height h, then Work done by lifting force F

${W_1} = Fh\cos 0 = Fh$

Again, work done by the force of gravity mg.

${W_2} = (mg)\,(h)\,\cos {180^o}$

${W_2} = - mgh$

Example 2

If a body is pulled on a rough horizontal surface through a displacement S.

Work done by normal reaction and gravity

${W_N} = NS\cos {90^o} = 0$ ;

Work done by pulling force F.

${W_F} = FS\cos {0^o} = FS$ ;

Work done by frictional force

${W_f} = f\,s\cos {180^o} = - f\,s$ ;

How is the physics definition of work done different from what people usually think about work?

Suppose a person a holding a box of mass m on his head for some time (say for two hours). Here, forces acting on the box are (i) mg force along the vertically downward direction and (ii) normal reaction force exerted by his head along the vertically upward direction. But the displacement of the points of application of forces is zero. So, according to the definition of work done in physics, work done by forces equals zero. One can say that the person holding a box on his head is doing work in daily life.

Calculation of Work done.

There are three approaches to calculating work done.

Algebraic approach
Graphical approach
Calculus approach

The algebraic approach has limited use, whereas the calculus approach is applicable in every situation.

The main advantage of the algebraic approach is that one need not know any calculus to find work done.

The algebraic approach of work done.

When a constant force F acts on a particle, and the particle moves through a displacement S, then the force is said to do work W on the particle $W = {\bf{F}} \cdot {\bf{S}}$

The scalar product of F and S can be evaluated as $W = {\bf{F}} \cdot {\bf{S}} = FS\cos \theta$

Here, F is the magnitude of force F,

S is the magnitude of displacement S and

$\theta$ is the smaller angle between F & S.

$W = FS\cos \theta = F(S\cos \theta )$

= magnitude of force × component of displacement along the direction of the force.

Or, $W = (F\cos \theta ) \cdot S$

= component of the force in the direction of displacement × magnitude of displacement

Example 1

A body of mass m is sliding down on a smooth inclined plane of inclination $\theta$ . If L is the length of the inclined plane, then work done by the gravitational force is

${W_{gravity}} = FS = (mg\sin \theta )\,L$

Here $mg\sin \theta$ is the component of the force of gravity along with displacement L.

Example 2

Work done in pulling the bob of mass m of a simple pendulum of length L through an angle $\theta$ to the vertical using horizontal external force F.

Work done by the gravitational force

${W_{gravity}} = mgh\cos {180^o}$

$= - mgL(1 - \cos \theta )$

$\cos \theta = \frac{{L - h}}{L} = 1 - \frac{h}{L}$

$\frac{h}{L} = 1 - \cos \theta$

$h = L(1 - \cos \theta )$

$\sin \theta = \frac{{AC}}{L}$

$AC = L\sin \theta$

Work done by the horizontal external force

${W_F} = F(AC) = F \cdot L\sin \theta$

Similarly, work done by tension developed in the string equals zero.

$W = {\bf{T}} \cdot {\bf{S}} = TS\cos {90^o} = 0$

Graphical representation of work done

The area enclosed by the F-S curve and displacement axis gives the amount of work done by the force.

Work = FS = Area (ABCO)

Work done by a variable force

For small displacement (dx), the work done will be the area of the strip of width dx.

dA = F(x) (dx) = dW

$W = \int {dW} = \int_{\,{x_i}}^{\,{x_f}} {F(x)} \,dx$

If the area enclosed above the x-axis, work done is +ve, and the area enclosed below the x-axis work done is –ve.

E.g., Work done by the spring force graphically (without calculus)

If x is the displacement of the free end of the spring from its equilibrium position, then the spring force can be written as ${F_{spring}} = - kx$

Here k is the spring constant.

The negative sign signifies that the force always opposes the expansion (x > 0) and compression (x < 0) of the spring. In other words, the force tends to restore the system to its equilibrium position.

So, the area enclosed under ${F_{sp}}$ vs x curve and displacement axis.

Area = Area of trapezium $= \frac{1}{2}(k{x_f} + k{x_i}) \times ({x_f} - {x_i})$ .

Here, $(k{x_f} + k{x_i})$ represents sum of parallel sides and $({x_f} - {x_i})$ represents separation between parallel sides.

$Area = \frac{k}{2}({x_f} + {x_i})({x_f} - {x_i}) = \frac{k}{2}(x_f^2 - x_i^2)$

But area enclosed below the x-axis is said to be negative.

Work done by the spring $= - \frac{k}{2}(x_f^2 - x_i^2)$ .

Calculus approach of work done

When the magnitude and direction of a force vary in three dimensions, it can be expressed as a function of the position vector F(r) or in terms of coordinates F(x, y, z). The work done by such a force in an infinitesimal displacement dS is

$dW = {\bf{F}} \cdot {\bf{dS}}$

${(W)_{1 \to 2}} = \int_{\,1}^{\,2} {{\bf{F}} \cdot {\bf{dS}}}$

In terms of rectangular components

${\bf{F}} = {F_x}\hat i + {F_y}\hat j + {F_z}\hat k$ and ${\bf{dS}} = ({d_x})\hat i + ({d_y})\hat j + ({d_z})\hat k$

Therefore,

${W_{1 \to 2}} = \int_{\,1}^{\,2} {\left( {{F_x}\hat i + {F_y}\hat j + {F_z}\hat k} \right) \cdot \left( {{d_x}\hat i + {d_y}\hat j + {d_z}\hat k} \right)}$

${W_{1 \to 2}} = \int_{\,{x_1}}^{\,{x_2}} {{F_x}\,dx} + \int_{\,{y_1}}^{\,{y_2}} {{F_y}\,dy} + \int_{\,{z_1}}^{\,{z_2}} {{F_z}\,dz}$

Example

A force ${\bf{F}} = (2x\hat i + 2\hat j + 3{z^2}\hat k)\,N$ is acting on a particle. Find the work done by the force in displacing the body from (1, 2, 3)m to (3, 6, 1)m.

Ans. Work done by force F can be written in the form-

$W = \int_{\,{x_i}}^{\,{x_f}} {{F_x}\,dx} + \int_{\,{y_i}}^{\,{y_f}} {{F_y}\,dy} + \int_{\,{z_i}}^{\,{z_f}} {{F_z}\,dz}$

$= \int_{\,1}^{\,3} {2x\,(dx)} + \int_{\,2}^{\,6} {2\,(dy)} + \int_{\,3}^{\,1} {3{z^2}\,(dz)}$

$= 2\left[ {\frac{{{x^2}}}{2}} \right]_1^3 + 2\left[ y \right]_2^6 + 3\left[ {\frac{{{x^3}}}{3}} \right]_3^1$

= – 10 J

Here (1, 2, 3)m is the initial position co-ordinate of particle, and (3, 6, 1)m is the final position co-ordinates of particle.

Work done by the spring force (with calculus)

Work done by the spring force for a displacement from ${x_i}$ to ${x_f}$ is given by

${W_{sp}} = \int_{\,{x_i}}^{\,{x_f}} {{F_{sp}}(dx)} = \int {( - kx)(dx)} = - \frac{k}{2}[{x^2}]_{x{ & _i}}^{{x_f}}$

${W_{sp}} = - \frac{k}{2}(x_f^2 - x_i^2)$

Some more work done examples in physics

Work done by friction

Ex.: A block of mass m = 250 gram. Slides down an inclined plane of inclination 37^o with a uniform speed. Find the work done by the friction as the block slides through 1.0 m.

Sol.: In our question, the body is sliding down with uniform speed. Therefore, the net force acting on the block along the incline will be zero.

Since, ${a_x} = 0\,;\,\sum {{F_x}} = 0\,;\,{f_k} = mg\sin \theta$

Now, work done by the friction force ; $W = {{\bf{f}}_{\bf{k}}} \cdot {\bf{S}} = {f_k}S\cos {180^o} = - {f_k}S$

${W_{{\rm{frictional force}}}} = - mg\sin \theta \cdot S$

$= - \left( {\frac{1}{4}kg} \right)(10\,m{\rm{/}}{s^2}) \times (\sin {37^o}) \times (1\,m) = - 1.5\,Joule$

Work done by the gravity (horizontal surface)

Ex.: A boy weighing 200 N is to skid through 20 m on a straight road slowly. Find work done by gravity.

Solⁿ:- Here angle between the force of gravity and displacement is $\frac{\pi }{2}$ ; therefore, work is done by the gravity.

${W_{gravity}} = m{\bf{g}} \cdot {\bf{S}} = (mg)(S)\cos {90^o} = 0$

Hence, work done by gravity in this situation will be zero.

Work done by gravity on an inclined plane

Example

Suppose a weight 5N is moved up a frictionless inclined plane from R to Q. What is work done by the gravity in Joule.

Solution:- In given figure, angle made by mg and S is $(\cos {90^o} + \theta )$ .

Therefore, work done by the gravity,

$W = mgs\cos ({90^o} + \theta )$

$= (5N)\,(5m)\,( - \sin \theta )$

$= (5N)\,(m)\,\left( { - \frac{3}{{}}} \right) = - 15\,J$

Work done in a circular motion

In the given figure, a simple pendulum of length l is hanging, one end is kept fixed, and mass m is attached at the lower end. Mass m is projected with some speed ${v_l}$ to that it describes a complete circle. Find work done by gravity as bob of mass m crosses through the highest point of its trajectory.

Ans.:- Work done by gravity is

${W_g} = m{\bf{g}} \cdot {\bf{S}} = mgs\cos {180^o}$

${W_g} = mg(2l)( - 1) = - 2mgl$

Work done by the electric field (without calculus)

Ex.: Suppose a particle of mass m and charge q is thrown at a speed u against the direction of uniform electric field E. How much distance will it travel before coming to rest momentarily.

Solution:- we can write the electric force experienced by charge q as $E = \frac{F}{q}$

$F = qE \Rightarrow ma = qE$

$a = \frac{{qE}}{m}$

Since the magnitude of the acceleration is constant, the projected particle is moving along a straight line.

Therefore, we can use, ${v^2} = {u^2} + 2as$

$0 = {u^2} + 2\left( { - \frac{{qE}}{m}} \right)(s)$ (-ve sign shows that velocity and acceleration have opposite sign)

$\frac{{2qE}}{m}s = {u^2}$

$s = \frac{{m{u^2}}}{{2qE}}$

Now, we can find the work done by the electric field as

$W = {{\bf{F}}_{el}} \cdot s$

${W_{el}} = qE \cdot s\cos {180^o}$

${W_{el}} = \cdot \frac{{m{u^2}}}{{2}}( - 1)$

${W_{el}} = \left( { - \frac{{m{u^2}}}{2}} \right)$ .

Work done to move a charge by external force and work done by an electric field. (example with calculus)

Let a positive charge Q is kept fixed, and a negative charge (-q) is moved away slowly with the help of external force. We have to calculate work done by external force as (-q) charge particle changes its position from $r = {r_i}$ to $r = {r_f}$ . Also, calculate work done by the electric field.

Since charge particle (-q) slowly moving along x-axis; therefore, $\left| {{{\bf{F}}_{ext}}} \right| = \parallel \left| {{{\bf{F}}_{el}}} \right|$

Therefore, work done by external agent $d{W_{ext}} = {{\bf{F}}_{ext}} \cdot {\bf{d}}r = {F_{el}} \cdot (dr)\cos {0^o} = \frac{{kQq}}{{{r^2}}}(dr)$

${W_{ext}} = kQq\int\limits_{{r_i}}^{{r_f}} {{r^{ - 2}}(dr)}$

${W_{ext}} = kQq\left[ {\frac{{{r^{ - 2 + 1}}}}{{ - 1}}} \right]_{{r_i}}^{{r_f}}$

${W_{ext}} = - kQq\left[ {\frac{1}{{{r_f}}} - \frac{1}{{{r_i}}}} \right]$

$\Rightarrow {W_{ext}} = kQq\left[ {\frac{1}{{{r_i}}} - \frac{1}{{{r_f}}}} \right]$ .

Again, as we know from the work-energy theorem for a system of particles ;

${({k_f})_{sy}} - {({k_i})_{sy}} = {W_{{\mathop{\rm int}} }} + {W_{ext}}$

$\Rightarrow 0 = {W_{el}} + {W_{ext}}$

Therefore work done by the electric field is equal to –ve of work done by external force in this situation.

Work done in Thermodynamics

Thermodynamics is a branch of science that deals with the exchange of heat energy between bodies and the conversion of heat energy into mechanical energy and vice-versa.

When a gas expands, then for an infinitesimally small change in volume dV, small work done dW = PdV.

Here P is almost constant.

Here,

dW = PdV = Area of narrow strip as shown in figure.

If the volume changes from ${V_1}$ to ${V_2}$ then the total work done by the system,

$W = \int\limits_{{V_1}}^{{V_2}} {P(dV)}$ = Area under P-V curve

The area under the P–V graph equals external work done during the process.

Examples

(i) A refrigerator works isothermally. A set of changes occur in the refrigerator’s mechanism, but the temperature inside always remains constant. Here, the heat energy is removed and transmitted to the surrounding environment.

(ii) Another example of thermodynamics is the heat pump. The heat is either removed from the house and dumped outside, or the heat is brought inside the house from outside to warm the house. In either case, the purpose is to keep the house at the desired temperature.

Work done by the system or work done on the system

If the work is done by the system, the volume of gas increases, then the work done is said to be positive.

$\left( {{V_f} - {V_i}} \right) > 0$

or, $dV > 0$

i.e., $dW > 0$

If the work is done on the system, then the volume of gas decreases.

$\left( {{V_f} - {V_i}} \right) < 0$

or, $dV < 0$

i.e., $dW < 0$

Work done in case of Isothermal process

In this process, the pressure and volume of gas change, but the gas temperature remains constant.

Here, dT = 0, dU = 0

Work done by the gas in Isothermal process

$\begin{array}{*{20}{c}} {W = }\\ {W = } \end{array}\left\{ {\begin{array}{*{20}{c}} {P(dV)}\\ {\,\,\frac{{nRT}}{V}(dV)} \end{array}} \right.$

Using PV = nRT. (Ideal gas equation).

$W = nRT\int\limits_{{V_i}}^{{V_f}} {\frac{1}{V}\left( {dv} \right)}$

$W = nRT\left[ {lnV} \right]_{{V_i}}^{{V_f}}$

$W = nRT\left( {ln\frac{{{V_f}}}{{{V_i}}}} \right)$

or, $W = nRT \cdot \left( {2.303\;{{\log }_{10}}\left( {\frac{{{V_f}}}{{{V_i}}}} \right)} \right)$

or, $W = nRT \cdot \left( {2.303\;{{\log }_{10}}\frac{{{P_f}}}{{{P_i}}}} \right)$

Here, W is work done during the isothermal change at temperature T for n moles of gas.

An Isothermal process is ideal. In nature, no process is perfectly isothermal. But we can say that melting ice and boiling water are approximately isothermal.

Work done in case of Adiabatic process

The pressure, volume, and temperature of a gas in an adiabatic process change, but total heat remain constant.

Q = Const, dQ = 0

There should not be any exchange of heat between the system and surroundings. This is a quick process and the internal energy changes as temperature changes.

In the adiabatic process, P, V, and T are related as

(i) $P{V^\gamma } = {\rm{constant}}$ (ii) $T{V^{\gamma - 1}} = {\rm{constant}}$ (iii) ${P^{1 - \gamma }}{T^\gamma } = {\rm{constant}}$

The work done by the system during the adiabatic expansion is

${W_{ad}} = \int\limits_{{V_i}}^{{V_f}} {P\left( {dV} \right)}$

${W_{ad}} = \int\limits_{{V_i}}^{{V_f}} {\frac{C}{{{V^\gamma }}}\left( {dV} \right)}$

Here, C is a constant and we can write ${P_i}{V_i}^\gamma = {P_f}{V_f}^\gamma = C$

${W_{ad}} = C\int\limits_{{V_i}}^{{V_f}} {{V^{ - \gamma }}\left( {dV} \right) = C\left( {\frac{{{V^{ - \gamma + 1}}}}{{ - \gamma + 1}}} \right)_{{V_i}}^{{V_f}}}$

${W_{ad}} = C\left[ {\frac{{{V^{1 - \gamma }}}}{{1 - \gamma }}} \right]_{{V_i}}^{{V_f}} = \frac{{CV_f^{1 - \gamma } - CV_i^{1 - \gamma }}}{{1 - \gamma }}$

${W_{ad}} = \left[ {\frac{{{P_f}V_f^\gamma V_f^{1 - \gamma } - {P_i}V_i^\gamma V_i^{1 - \gamma }}}{{\left( {1 - \gamma } \right)}}} \right]$

${W_{ad}} = \frac{{{P_i}{V_i} - {P_f}{V_f}}}{{\left( {\gamma - 1} \right)}}$

Here $\gamma$ is the ratio of molar heat capacity of gas at constant pressure and molar heat capacity at constant volume.

It takes place in a non-conducting vessel. Hence no exchange of heat takes place between the system and its surroundings.

An adiabatic expansion causes cooling and compression causing heating (dV = – dW).

Example of adiabatic process

Sudden bursting of a tube of a bicycle tire, propagation of sound in gases.

Work done in case of Isobaric process

It is a process in which the pressure of the system remains constant,

i.e., $\Delta P = 0 \Rightarrow {P_i} = {P_f}$

So, Work done in an Isobaric process is

$W = \int\limits_{{V_i}}^{{V_f}} {P\left( {dV} \right) = P\int\limits_{{V_i}}^{{V_f}} {\left( {dV} \right)} }$

${W_{\scriptstyleIsobaric\hfill\atop \scriptstyleprocess\hfill}} = P\left( {{V_f} - {V_i}} \right)$

Work done in case of Free expansion

The figure shows an insulated cylinder divided into two parts by a thin massless fixed piston. The volume of the lift compartment gets filled with an ideal gas, and the right compartment is a vacuum. If we release the piston, gas fills the whole space of the cylinder rapidly. In this expansion, no heat is supplied to the gas as the walls are insulated.

$\quad \Delta Q = 0,$ as the piston is fixed, no work is done by the gas, $\Delta W = 0$

Therefore, internal energy remains constant.

$\Delta U = 0\,;\,\,T:{\rm{constant}}$

We call such an expansion, free expansion.

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