Physics > Work Energy and Power > 2.0 Work done by a constant force
Work Energy and Power
1.0 Introduction
2.0 Work done by a constant force
3.0 Spring Force
4.0 Conservative & Non-conservative forces
5.0 Kinetic Energy $(K)$
6.0 Potential energy $\left( {\Delta U} \right)$
6.1 Potential energy $\left( {\Delta U} \right)$ is negative of the work done by conservative forces.
6.2 Types of potential energy
6.3 Law of conservation of mechanical energy
7.0 Work energy theorem
8.0 Power
9.0 Types of equilibrium
10.0 Work done by a distributed mass
2.1 Properties of work done
6.2 Types of potential energy
6.3 Law of conservation of mechanical energy
- Work done is a scalar quantity as it is a dot product of two vector quantities $\overrightarrow F$ and $\overrightarrow s$.
- Work done can be positive, negative or zero. It depends on the angle $\theta$ between the force $(F)$ and the displacement $(s)$.
- When $\theta < 90^\circ $: work done $(W)$ is positive
- When $\theta = 90^\circ $ work done $(W)$ is zero
- When $\theta > 90^\circ $: work done $(W)$ is negative
- Work done depends on the frame of reference
A. Inertial frame of reference
- According to the observer A, the block has displacement $s$. Therefore the work done by the force is, $$W = \overrightarrow F .\overrightarrow s $$
- In this case, the train is moving with constant velocity and $d$ & $s$ is the displacement of the train and the block respectively. According to the observer A, the block has displacement $s$. Therefore the work done by the force is, $$W = \overrightarrow F .\overrightarrow s $$
- According to the observer B, the block has displacement $(s+d)$. Therefore the work done by the force is, $$W = \overrightarrow F .\left( {\overrightarrow s + \overrightarrow d } \right)$$
B. Non-inertial frame of reference
- In this case the train is moving with constant acceleration $a$ and $d$ & $s$ is the displacement of the train and the block respectively. According to the observer A, the block has displacement $s$. Therefore the work done by the force is,$$W = \overrightarrow F .\overrightarrow s $$
- According to the observer B, the block has displacement $(s+d)$ and the net force acting on the block in the horizontal direction is $(F-ma)$. Therefore the net work done is, $$W = \left( {\overrightarrow F - m\overrightarrow a } \right).\left( {\overrightarrow s + \overrightarrow d } \right)$$
- According to the observer B, the block has displacement $(s+d)$. Therefore the work done by the force $\overrightarrow F $ is, $$W = \overrightarrow F .(\overrightarrow s + \overrightarrow d )$$
Question 3. A body is displaced from point $P$ $\left( {4\widehat i + 2\widehat j + 3\widehat k} \right)$ to $Q$ $\left( { - 3\widehat i - 2\widehat j + 5\widehat k} \right)$. Find the work done by a constant force $\overrightarrow F$ $\left( { 5\widehat i + 3\widehat j + 8\widehat k} \right)$.
Solution: Displacement of a body under the constant force $\overrightarrow F$ is ${\overrightarrow s _{PQ}}$.
$$\begin{equation} \begin{aligned} {\overrightarrow s _{PQ}} = {\overrightarrow s _P} - {\overrightarrow s _Q} \\ {\overrightarrow s _{PQ}} = \left( {4\widehat i + 2\widehat j + 3\widehat k} \right) - \left( { - 3\widehat i - 2\widehat j + 5\widehat k} \right) \\ {\overrightarrow s _{PQ}} = \left( {7\widehat i + 4\widehat j - 2\widehat k} \right) \\\end{aligned} \end{equation} $$
We know, work done $(W)$ is given by, $$\begin{equation} \begin{aligned} W = \overrightarrow F .\overrightarrow s \\ W = \left( {5\widehat i + 3\widehat j + 8\widehat k} \right).\left( {7\widehat i + 4\widehat j - 2\widehat k} \right) \\ W = (35 + 12 - 16) \\ W = 31J \\\end{aligned} \end{equation} $$
Question 4. A truck is moving with constant velocity $v=10 m/s$. A constant force $F=2 N$ relative to the truck is applied on a block of mass $m=1 kg$ which is initially at rest as shown in the figure. Find the work done on the block after 3 seconds as seen by,
- Observer 1
- Observer 2
Solution: Displacement of truck $(s_1)$ in 3 seconds $=10\times 3=30m$
Acceleration of block due to constant horizontal force $F$ is $a=2m/s^2$
Displacement of block relative to truck $(s_2)$ in 3 seconds $=\frac{1}{2}at^2=\frac{1}{2}\times 2\times 9=9m$
1.Displacement of block as observed by observer 1, $=9m$
Work done $(W_1)$ is given by, $$\begin{equation} \begin{aligned} {W_1} = \overrightarrow F .\overrightarrow s \\ {W_1} = 2 \times 9 \\ {W_1} = 18J \\\end{aligned} \end{equation} $$
2.Displacement of block as observed by observer 2 = Displacement of truck + Displacement of block relative to the truck $$\begin{equation} \begin{aligned} \Rightarrow (30 + 9)m \\ \Rightarrow 39m \\\end{aligned} \end{equation} $$
Work done $(W_2)$ is given by, $$\begin{equation} \begin{aligned} {W_2} = \overrightarrow F .\overrightarrow s \\ {W_2} = 2 \times 39 \\ {W_2} = 78J \\\end{aligned} \end{equation} $$
