Question

A particle moves with a velocity $\overrightarrow{v}=(6\widehat{i}-4\widehat{j}+3k)m{{s}^{-1}}$ under the influence of a constant force $\overrightarrow{F}=(20\overset\frown{i}+15\widehat{j}-5\widehat{k})N$. The instantaneous power applied to the particle is –

& \text{A) 35J}{{\text{s}}^{-1}} \\\ & \text{B) 45J}{{\text{s}}^{-1}} \\\ & \text{C) 25J}{{\text{s}}^{-1}} \\\ & \text{D) 195J}{{\text{s}}^{-1}} \\\ \end{aligned}$$

Answer 1

The particle is moving with a velocity under a constant force. We can find the power used up by the particle from this force directly using the dot product of the two quantities force and the velocity. The power will then turn out to be scalar quantity unlike the force and velocity.

Complete answer:
The power applied on a particle or mass is dependent on the applied constant force and the velocity attained by the mass during the time of application of force. We can derive this conclusion from the basic relations between energy and power.
The power is defined as the rate of change of energy. Whereas, the energy is the force applied in moving an object by unit displacement. So, when we relate these quantities, we get the power applied as –

& P=\dfrac{E}{t} \\\ & \text{also,} \\\ & E=F.s \\\ & \Rightarrow \text{ }P=\dfrac{F.s}{t} \\\ & \text{but,} \\\ & v=\dfrac{s}{t} \\\ & \Rightarrow \text{ }P=F.v \\\ \end{aligned}$$ We are given the vector forms of the velocity and force, from which we can find the power applied on the particle as – $$\begin{aligned} & \text{Given,} \\\ & \overrightarrow{F}=(20\overset\frown{i}+15\widehat{j}-5\widehat{k})N \\\ & \overrightarrow{v}=(6\widehat{i}-4\widehat{j}+3k)m{{s}^{-1}} \\\ & \therefore \text{ }P=F.v \\\ & \\\ \end{aligned}$$ $$\begin{aligned} &\Rightarrow\text{}P=\text{}(20\overset\frown{i}+15\widehat{j}-5\widehat{k}).(6\widehat{i}-4\widehat{j}+3k) \\\ & \Rightarrow \text{ }P=120-60-15 \\\ & \Rightarrow \text{ }P=45J{{s}^{-1}} \\\ \end{aligned}$$ The power applied is $$45J{{s}^{-1}}$$. **So, the correct answer is “Option B”.** **Additional Information:** The work done or energy consumed by applying a force is given by the dot or scalar product of the force and the displacement. A perpendicular force thus results in zero work done. **Note:** The power applied by the force on a particle moving with a velocity due to this applied force can be zero, if the force applied is perpendicular to the velocity, the force component along the horizontal component is zero or when the particle doesn’t move at all with the applied force.