Question

A force of $50N$ acts on a body and moves it at a distance of $4m$ on horizontal surface. Calculate the work done if the direction of force is at an angle of $60^\circ$ to the horizontal surface. What will be the value of work done if the angle is $30^\circ$ instead of $60^\circ$ ?

Answer 1

In order to solve this question we need to understand force and work done. Force is defined as push or pull on an object which causes the state change of the object, either bringing it to rest or in motion. Work done is defined as the amount of force in a direction of displacement of an object that causes a unit displacement in an object. By doing work, energy is being stored in an object in the form of potential or kinetic energy.

Complete step by step answer:
According to question, magnitude of force is, $\left| {\vec F} \right| = 50\,N$.
And displacement in the body due to force is, $d = 4\,m$.
Also the angle from horizontal at which force acts is, $\theta = 60^\circ$
Let the work done by a body be “W”. Then the work done is defined as, $W = \vec F.\vec d$.
Removing the dot product we have,
$W = Fd\cos \theta$
Putting values we get,
$W = 50 \times 4 \times \cos (60)$
$\Rightarrow W = 200 \times \dfrac{1}{2}$ Since $\cos (60) \\\ \Rightarrow W= \dfrac{1}{2}$
$\Rightarrow W = 100\,J$
So the work done when $\theta = 60^\circ$ is, $W = 100J$.

If the angle from horizontal that force make be,
$\varphi = 30^\circ$
Then the work done is defined as,
$W = \vec F.\vec d$
Removing the dot product we have,
$W = Fd\cos \varphi$
Putting values we get,
$W = 50 \times 4 \times \cos (30)$
$\Rightarrow W = 200 \times \dfrac{{\sqrt 3 }}{2}$ Since $\cos (60) = \dfrac{{\sqrt 3 }}{2}$.
$\therefore W = 100\sqrt 3 J$

So the work done when $\varphi = 30^\circ$ is, $W = 100\sqrt 3 \,J$.

Note: It should be remembered that work done can be positive, negative or zero. If the force is acting in direction of displacement then work done is positive, if force in opposite direction of displacement the work done is negative and if the force acts in direction perpendicular to displacement then work done is zero.