WHAT IS WORK?

Work refers to the force and the displacement over which it acts. Work is done by a force on the body over a certain displacement. Consider a constant force \(\vec{F}\) acting on an object of mass \(m\). The object undergoes a displacement \(\vec{d}\) in the positive \(x\)-direction as shown in Figure 6.1.

The work done by the force is defined to be the product of the component of the force in the direction of the displacement and the magnitude of this displacement. Thus
\(W= \vec{F} \vec{d} = (F \cos \theta) d  \dots (6.4)\)

We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, the force you exert on the wall does no work. \(\text { The SI unit of work is Joule }(\mathrm{J}) \text {, }\) and its \(\text { dimension is, }\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\).

No work is done if:

• the displacement is zero as seen in the example above. A weightlifter holding a 150 kg mass steadily on his shoulder for \(30 \mathrm{~s}\) does no work on the load during this time.
• the force is zero. A block moving on a smooth horizontal table is not acted upon by a horizontal force (since there is no friction), but may undergo a large displacement.
• the force and displacement are mutually perpendicular. This is so since, for \(\theta=\pi / 2 \mathrm{rad}\) \(\left(=90^{\circ}\right), \cos (\pi / 2)=0\). For the block moving on a smooth horizontal table, the gravitational force \(m g\) does no work since it acts at right angles to the displacement. If we assume that the moon’s orbit around the Earth is perfectly circular then the Earth’s gravitational force does no work. The moon’s instantaneous displacement is tangential while the earth’s force is radially inwards and \(\theta=\pi / 2\).
• Work can be both positive and negative. If \(\theta\) is between \(0^{\circ}\) and \(90^{\circ}, \cos \theta\) in Eq. (6.4) is positive. If \(\theta\) is between \(90^{\circ}\) and \(180^{\circ}, \cos \theta\) is negative. In many examples, the frictional force opposes displacement and \(\theta=180^{\circ}\). Then the work done by friction is negative \(\left(\cos 180^{\circ}=-1\right)\).

Calculation Of Work Done

The work done by a force on a particle during a displacement has been defined as
\(
W=\int \vec{F} \cdot d \vec{r}
\)

Case 1: Constant Force

Suppose, the force is constant (in direction and magnitude) during the displacement. Then \(W=\int \vec{F} \cdot d \vec{r}=\vec{F} \cdot \int d \vec{r}=\vec{F} \cdot \vec{r}\), where \(\vec{r}\) is the total displacement of the particle during which the work is calculated. If \(\theta\) be the angle between the constant force \(\vec{F}\) and the displacement \(\vec{r}\), the work is
\(
W=\operatorname{Fr} \cos \theta .
\)
In particular, if the displacement is along the force, as is the case with a freely and vertically falling particle, \(\theta=0\) and \(W=F r\).

The force of gravity \((\mathrm{mg})\) is constant in magnitude and direction if the particle moves near the surface of the earth. Suppose a particle moves from \(A\) to \(B\) along some curve and that \(\overrightarrow{A B}\) makes an angle \(\theta\) with the vertical as shown in the figure below. The work done by the force of gravity during the transit from \(A\) to \(B\) is
\(
W=m g(A B) \cos \theta=m g h,
\)
where \(h\) is the height descended by the particle. If a particle ascends a height \(h\), the work done by the force of gravity is \(-m g h\).

If the particle goes from point \(A\) to point \(B\) along some other curve, the work done by the force of gravity is again \(\mathrm{mgh}\). We see that the work done by a constant force in going from \(A\) to \(B\) depends only on the positions of \(A\) and \(B\) and not on the actual path taken. In case of gravity, the work is weight \(\mathrm{mg}\) times the height descended. If a particle starts from \(A\) and reaches to the same point \(A\) after some time, the work done by gravity during this round trip is zero, as the height descended is zero. We shall encounter other forces having this property.

Case-2: Spring Force

Consider the situation shown in the figure below. One end of a spring is attached to a fixed vertical support and the other end to a block which can move on a horizontal table. Let \(x=0\) denote the position of the block when the spring is in its natural length. We shall calculate the work done on the block by the spring force as the block moves from \(x=0\) to \(x=x_1\).

The force on the block is \(k\) times the elongation of the spring. But the elongation changes as the block moves and so does the force. We cannot take \(\vec{F}\) out of the integration \(\int \vec{F} \cdot d \vec{r}\). We have to write the work done during a small interval in which the block moves from \(x\) to \(x+d x\). The force in this interval is \(k x\) and the displacement is \(d x\). The force and the displacement are opposite in direction.
So, \(\quad \vec{F} \cdot d \vec{r}=-F d x=-k x d x\)
during this interval. The total work done as the block is displaced from \(x=0\) to \(x=x_1\) is
\(
W=\int_0^{x_1}-k x d x=\left[-\frac{1}{2} k x^2\right]_0^{x_1}=-\frac{1}{2} k x_1^2 .
\)
If the block moves from \(x=x_1\) to \(x=x_2\), the limits of integration are \(x_1\) and \(x_2\) and the work done is
\(
W=\left(\frac{1}{2} k x_1^2-\frac{1}{2} k x_2^2\right) \text {. }
\)
Note that if the block is displaced from \(x_1\) to \(x_2\) and brought back to \(x=x_1\), the work done by the spring force is zero. The work done during the return journey is negative of the work during the onward journey. The net work done by the spring force in a round trip is zero.

Three positions of a spring are shown in the figure below. In (i) the spring is in its natural length, in (ii) it is compressed by an amount \(x\) and in (iii) it is elongated by an amount \(x\). 

Work done by the spring force on the block in various situations is shown in the following table.

\(
\begin{aligned}
&\text { Table } 6.1\\
&\begin{array}{llccc}
\hline \begin{array}{l}
\text { Initial state of } \\
\text { the spring }
\end{array} & \begin{array}{l}
\text { Final state of } \\
\text { the spring }
\end{array} & x_1 & x_2 & W \\
\hline \text { Natural } & \text { Compressed } & 0 & -x & -\frac{1}{2} k x^2 \\
\text { Natural } & \text { Elongated } & 0 & x & -\frac{1}{2} k x^2 \\
\text { Elongated } & \text { Natural } & x & 0 & \frac{1}{2} k x^2 \\
\text { Compressed } & \text { Natural } & -x & 0 & \frac{1}{2} k x^2 \\
\text { Elongated } & \text { Compressed } & x & -x & 0 \\
\text { Compressed } & \text { Elongated } & -x & x & 0 \\
\hline
\end{array}
\end{aligned}
\)

Force Perpendicular to Velocity

Suppose \(\vec{F} \perp \vec{v}\) for all the time. Then \(\vec{F} \cdot d \vec{r}=\vec{F} \cdot \vec{v} d t\) is zero in any small interval and the work done by this force is zero.

For example, if a particle is fastened to the end of a string and is whirled in a circular path, the tension is always perpendicular to the velocity of the particle and hence the work done by the tension is zero in circular motion.

The following three cases occur quite frequently:
(a) The force is perpendicular to the velocity at all the instants. The work done by the force is then zero.
(b) The force is constant (both in magnitude and direction). The work done by the force is \(W=F d \cos \theta\), where \(F\) and \(d\) are magnitudes of the force and the displacement and \(\theta\) is the angle between them. The amount of work done depends only on the end positions and not on the intermediate path. The work in a round trip is zero. Force of gravity on the bodies near the earth’s surface is an example.

The work done due to the force of gravity on a particle of mass \(m\) is \(m g h\), where \(h\) is the vertical height ‘descended’ by the particle.
(c) The force is \(F=-k x\) as is the case with an elastic spring. The magnitude of the work done by the force during a displacement \(x\) from or to its natural position \((x=0)\) is \(\frac{1}{2} k x^2\). The work may be \(+\frac{1}{2} k x^2\) or \(-\frac{1}{2} k x^2\) depending on whether the force and the displacement are along the same or opposite directions.

Example 6.4: A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road? 

Solution: Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.
(a) The stopping force and the displacement make an angle of \(180^{\circ}\) ( \(\pi\) rad) with each other. Thus, work done by the road,
\(
\begin{aligned}
W_{r} &=F d \cos \theta \\
&=200 \times 10 \times \cos \pi \\
&=-2000 \mathrm{~J}
\end{aligned}
\)
It is this negative work that brings the cycle to a halt in accordance with the Work-Energy theorem.
(b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is \(200 \mathrm{~N}\). However, the road undergoes no displacement. Thus, work done by cycle on the road is zero.

Example 6.5: A spring of spring constant \(50 \mathrm{~N} / \mathrm{m}\) is compressed from its natural position through \(1 \mathrm{~cm}\). Find the work done by the spring force on the agency compressing the spring.

Answer: The magnitude of the work is
\(
\begin{aligned}
\frac{1}{2} k x^2 & =\frac{1}{2} \times(50 \mathrm{~N} / \mathrm{m}) \times(1 \mathrm{~cm})^2 \\
& =(25 \mathrm{~N} / \mathrm{m}) \times\left(1 \times 10^{-2} \mathrm{~m}\right)^2=2.5 \times 10^{-3} \mathrm{~J} .
\end{aligned}
\)
As the compressed spring will push the agency, the force will be opposite to the displacement of the point of
application and the work will be negative. Thus, the work done by the spring force is \(-2.5 \mathrm{~mJ}\).

Example 6.6: A particle of mass \(20 \mathrm{~g}\) is thrown vertically upwards with a speed of \(10 \mathrm{~m} / \mathrm{s}\). Find the work done by the force of gravity during the time the particle goes up.

Answer: Suppose the particle reaches a maximum height \(h\). As the velocity at the highest point is zero, we have
\(
\begin{aligned}
0 & =u^2-2 g h \\
\text { or, } \quad h & =\frac{u^2}{2 g} .
\end{aligned}
\)
or,
The work done by the force of gravity is
\(
\begin{aligned}
& -m g h=-m g \frac{u^2}{2 g}=-\frac{1}{2} m u^2 \\
& =-\frac{1}{2}(0.02 \mathrm{~kg}) \times(10 \mathrm{~m} / \mathrm{s})^2=-1.0 \mathrm{~J} .
\end{aligned}
\)