# Work

Thw work done by a force $\vec{F}$ in displacing an object by a displacement $\vec{S}$ is given by \begin{align} W=\vec{F}\cdot\vec{S}=FS\cos\theta. \end{align} The unit of work is Joule. Its dimensions are ML2T-2. The dimensions of work and torque are the same.

The work done by a force $\vec{F}$ is calculated by integrating over the path \begin{align} W=\int_\text{path}\vec{F}\cdot\mathrm{d}\vec{s}. \end{align} The work is equal to the area under the force-displacement curve.

## Problems from IIT JEE

Problem: A force $\vec{F}=3\hat\imath+c\hat\jmath+2\hat{k}$ acting on a particle causes a displacement $\vec{S}=-4\hat\imath+2\hat\jmath-3\hat{k}$. If the work done by the force is zero then the value of $c$ is

1. 0
2. 4
3. 9
4. 18

Solution: The work done by a force $\vec{F}=3\,\hat\imath+c\,\hat\jmath+2\,\hat{k}$ that causes a displacement $\vec{S}=-4\,\hat\imath+2\,\hat\jmath-3\,\hat{k}$ is given by $W=\vec{F}\cdot\vec{S}$. Since work done by the force is zero, we get \begin{align} W&=\vec{F}\cdot\vec{S}\\ &=-12+2c-6\\ &=0,\nonumber \end{align} which gives $c=9$. Note that the work done by a non-zero force that cause non-zero displacement is zero only if force is perpendicular to the displacement.

Problem (JEE Mains 2004): A force $F=(5\hat\imath + 3\hat\jmath + 2\hat{k})$ N is applied over a particle which displaces it from the origin to the point $\vec{r} = (2\hat\imath-\hat\jmath)$ m. The work done on the particle is

1. -7 J
2. 7 J
3. 10 J
4. 13 J

Solution: A constant force $F=(5\hat\imath + 3\hat\jmath + 2\hat{k})$ N displaces the particle by $\vec{s}=(2\hat\imath-\hat\jmath)$ m. The work done on the particle is, \begin{align} W&=\int_\text{path} \vec{F}\cdot\mathrm{d}\vec{s} \nonumber\\ &=\vec{F}\cdot \int_\text{path} \mathrm{d}\vec{s} \qquad\qquad (\because \vec{F} \text{ is const.}) \nonumber\\ &=\vec{F}\cdot\vec{s} \nonumber\\ &=(5\hat\imath + 3\hat\jmath + 2\hat{k})\cdot(2\hat\imath-\hat\jmath) \\ &=7\,\mathrm{J}.\nonumber \end{align} We encourage you to find $W$ by integrating over the path $(0,0)\to(2,0)\to(2,-1)$.

Problem (IIT JEE 2019): A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in the presence of a force \begin{align} \vec{F}=(\alpha y\,\hat\imath+2\alpha x\,\hat\jmath)\;\mathrm{N} \end{align} where $x$ and $y$ are in meter and $\alpha=-1$ N/m. The work done on the particle by this force $\vec{F}$ will be________Joule.