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Question: A force acts on a body and displaces it by a distance S in a direction at an angle \(\theta \) with ...

A force acts on a body and displaces it by a distance S in a direction at an angle θ\theta with the direction of force. What should be the value of θ\theta to get the maximum positive work ?

Explanation

Solution

In order to solve above problem first write the formula of work done in terms of θ\theta i.e.,
W=FScosθW = FS\cos \theta
Now use the concept that for maximum work done i.e., for maximum LHS, RHS also should be maximum.Hence, we will get a desired solution.

Complete step by step answer:
We know that if a body displaces by distance S on acting the force F, then work done by the force is
W=FScosθW = FS\cos \theta …..(1)
Where θ\theta is the angle between force F and displacement S.
Here, we have to calculate θ\theta for maximum work done W. So, for maximum work done RHS of equation 1 will also be maximum.
We know that the maximum value of cosθ\cos \theta is 1.
cosθ=1\Rightarrow \cos \theta = 1
cosθ=cos0\Rightarrow \cos \theta = \cos 0^\circ
θ=0\therefore \theta = 0^\circ

Thus, displacement of the body must be in the same direction as that of force to get the maximum positive work.

Note: In many problems, student may get confused between maximum and minimum values of sinθ\sin \theta and cosθ\cos \theta which are given as,
(cosθ)max=1{(\cos \theta )_{\max }} = 1 (sinθ)max=1{(\sin \theta )_{\max }} = 1
(cosθ)min=1{(\cos \theta )_{\min }} = - 1 (sinθ)min=0{(\sin \theta )_{\min }} = 0
And when θ\theta is very small then,
sinθθ,cosθ1\sin \theta \approx \theta ,\cos \theta \approx 1