A force acts on a body and displaces it by a distance S in a direction at an angle $\theta $ with the direction of force. What should be the value of $\theta $ to get the maximum positive work ?
Answer
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Hint:In order to solve above problem first write the formula of work done in terms of $\theta $ i.e.,
$W = 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 = FS\cos \theta $ …..(1)
Where $\theta $ is the angle between force F and displacement S.
Here, we have to calculate $\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 $\cos \theta $ is 1.
$\Rightarrow \cos \theta = 1$
$ \Rightarrow \cos \theta = \cos 0^\circ $
$ \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 $\sin \theta $ and $\cos \theta $ which are given as,
${(\cos \theta )_{\max }} = 1$ ${(\sin \theta )_{\max }} = 1$
${(\cos \theta )_{\min }} = - 1$ ${(\sin \theta )_{\min }} = 0$
And when $\theta $ is very small then,
$\sin \theta \approx \theta ,\cos \theta \approx 1$
$W = 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 = FS\cos \theta $ …..(1)
Where $\theta $ is the angle between force F and displacement S.
Here, we have to calculate $\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 $\cos \theta $ is 1.
$\Rightarrow \cos \theta = 1$
$ \Rightarrow \cos \theta = \cos 0^\circ $
$ \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 $\sin \theta $ and $\cos \theta $ which are given as,
${(\cos \theta )_{\max }} = 1$ ${(\sin \theta )_{\max }} = 1$
${(\cos \theta )_{\min }} = - 1$ ${(\sin \theta )_{\min }} = 0$
And when $\theta $ is very small then,
$\sin \theta \approx \theta ,\cos \theta \approx 1$
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