Question

Define work and write its units.

Answer 1

Hint: In Physics, work is calculated in terms of force and displacement. It is nothing but a process of energy transfer to an object by applying force. Work is said to be done only if there is a displacement caused due to the force. The SI unit of work is $joule(J)$.

Complete step-by-step answer:
In Physics, work is defined as the product of force and displacement. If an object is acted upon by a force, work done is nothing but the product of the magnitude of this force and the displacement of the object caused by the force. It is important that displacement occurs in the process. If there is no displacement, work done turns out to be zero. This is the reason why physicists say that a person trying to move a heavy object is doing no work. Although the person is applying force with a lot of energy, the heavy object restricts it to move itself from its position. Hence, there is no displacement happening and the work done is zero.
Displacement due to a force can happen in any direction. When force is applied on an object, a component of force can either be present in the same direction of displacement or in the opposite direction of displacement. Positive work is done if the direction of displacement is the same as the direction of force. Negative work is done if the direction of displacement is opposite to the direction of force.
Now, let us understand the equation of work.
$W=F\times d$
Here,
$F$ is the force acting on an object
$d$ is the displacement of the object caused by the force
$W$ is the work done by the force $F$ on the object
If the force is acting at an angle $\theta $ to the displacement, work done is given by
$W=Fd\cos \theta $
This equation can be taken as the general expression of work.
The SI unit of work is$joule(J)$. $1J$ is defined as the work done by a force of $1N$ to move an object through a distance of $1m$. Sometimes, $J$is also expressed as $newton-metre(N.m)$. Non-SI units of work include $erg$, the $foot-pound$, the $kilowatt-hour$, etc.

Note: If an object is dragged through a distance of $8m$on a horizontal surface by applying a force of $10N$, $\theta $ is $0$and $\cos \theta =1$. Work done in this case is:
$W=Fd\cos \theta =Fd(1)=Fd=10N\times 8m=80J$
This problem illustrates that $Fd\cos \theta $ can be used in any situation and can be taken as the general expression for work done.