Question

A particle of mass ‘m’ moves under the influence of the force $\overrightarrow{F}=a(\sin \omega t\widehat{i}+\cos \omega t\widehat{j})$, where a and $\omega$ are constants and ‘t’ is the time. The particle is initially at rest at the origin. The instantaneous power given to the particle is –

& \text{A) zero} \\\ & \text{B) }\dfrac{{{\text{a}}^{2}}\sin \omega t}{m\omega } \\\ & \text{C) }\dfrac{{{\text{a}}^{2}}\cos \omega t}{m\omega } \\\ & \text{D) }\dfrac{\text{ }{{\text{a}}^{2}}(\sin \omega t+\cos \omega t)}{m\omega } \\\ \end{aligned}$$

Answer 1

We are to find the power on a body which is under the influence of a given force. We can derive the relation between the force and power step-by-step to get the required answer for the problem. We know that the force and power are related to each other.

Complete answer:
We are given a force that is acting on a mass ‘m’ over a time ‘t’. We can use the known relations between force and other quantities which connect to the power to get the solution for the answer.
It is given that,
$\overrightarrow{F}=a(\sin \omega t\widehat{i}+\cos \omega t\widehat{j})$
We know that the force and mass give the acceleration as –

& A=\dfrac{F}{m} \\\ & \Rightarrow A=\dfrac{a(\sin \omega t\widehat{i}+\cos \omega t\widehat{j})}{m} \\\ & \therefore A=\dfrac{a}{m}(\sin \omega t\widehat{i}+\cos \omega t\widehat{j}) \\\ \end{aligned}$$ Now, we can relate the velocity of the particle at an instant ‘t’ to be derived by integrating the acceleration over the time ‘t’ as – $$\begin{aligned} & v=\int{A.dt} \\\ & \Rightarrow v=\int{\dfrac{a}{m}(\sin \omega t\widehat{i}+\cos \omega t\widehat{j}).dt} \\\ & \Rightarrow v=\dfrac{a}{m}\dfrac{1}{\omega }(-\cos \omega t\widehat{i}+\sin \omega t\widehat{j})+c \\\ & \Rightarrow v=\dfrac{a}{m\omega }(-\cos \omega t\widehat{i}+\sin \omega t\widehat{j})+c \\\ \end{aligned}$$ But we need to find the constant that adds to the velocity. Let us consider the initial conditions to be $$\begin{aligned} & t=0, \\\ & v=0 \\\ \end{aligned}$$ We can find the constant as – $$\begin{aligned} & 0=-\dfrac{a}{m\omega }\widehat{i}+c \\\ & \therefore c=\dfrac{a}{m\omega }\widehat{i} \\\ \end{aligned}$$ Therefore, the velocity will be given as – $$\Rightarrow v=\dfrac{a}{m\omega }(i-\cos \omega t\widehat{i}+\sin \omega t\widehat{j})$$ The instantaneous power of the particle when a force is applied and the particle moves with a velocity is given by – $$P=F.v$$ We can use the known data to find the instantaneous power using this relation as – $$\begin{aligned} & P=F.v \\\ & P=a(\sin \omega t\widehat{i}+\cos \omega t\widehat{j}).\dfrac{a}{m\omega }(i+\cos \omega t\widehat{i}-\sin \omega t\widehat{j}) \\\ & \Rightarrow P=\dfrac{{{a}^{2}}}{m\omega }(\sin \omega t+\sin \omega t\cos \omega t-co\operatorname{s}\omega t\sin \omega t) \\\ & \therefore P=\dfrac{{{a}^{2}}}{m\omega }\sin \omega t \\\ \end{aligned}$$ The instantaneous power is given as – $$P=\dfrac{{{a}^{2}}}{m\omega }\sin \omega t$$ **The correct answer is option B.** **Note:** We understand from the above example that the power is a scalar quantity, even though the force and velocity are vector quantities. The scalar product results in the scalar relation between the force, velocity and instantaneous power of the object under the force.