Question

A small motor is accelerating a mass $m$ cart from rest through a distance $d$ with force ${{F}_{0}}$​. If the output of force were doubled, how much more power would the motor be putting out in a certain distance?
$\begin{align}
  & A.{{P}_{0}} \\
 & B.\sqrt{2}{{P}_{0}} \\
 & C.2{{P}_{0}} \\
 & D.2\sqrt{2}{{P}_{0}} \\
 & E.4{{P}_{0}} \\
\end{align}$

Answer 1

Hint: Here it is mentioned that the cart is in rest. Therefore we can say that the initial velocity will be zero. Find the acceleration of the mass first. With these values of acceleration, using Newton's second equation of motion, the time taken for the travel is to be calculated. The work done by the motor is found. Then the power of the motor is determined by dividing the work done and the time taken.

Complete answer:
First of all let us mention what all are given in the question.
The cart is initially at rest so that the initial velocity can be taken as zero.
$u=0$
The distance up to which the cart moved is $d$ and the force applied by the motor will be ${{F}_{0}}$.
Therefore the acceleration of the motor will be given as,
$a=\dfrac{F}{m}$
Using the newton’s second law of motion,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Substitute the value of acceleration and distance travelled in it,
$d=ut+\dfrac{1}{2}\dfrac{{{F}_{0}}}{m}{{t}^{2}}$
As the initial velocity is zero,
$d=0+\dfrac{1}{2}\dfrac{{{F}_{0}}}{m}{{t}^{2}}$
The time taken to travel can be found by rearranging the equation,
$t=\sqrt{\dfrac{2md}{{{F}_{0}}}}$
The work done by the motor is given as,
$W={{F}_{0}}d$
Therefore the power required by the motor can be given as,
$P=\dfrac{W}{t}=\dfrac{{{F}_{0}}d}{\sqrt{\dfrac{2md}{{{F}_{0}}}}}$
From this we can see that the relation between the power and the force is,
\[P\propto {{F}_{0}}^{\dfrac{3}{2}}\]
Therefore we can compare using this,
\[\dfrac{{{P}'}}{{{P}_{0}}}=\dfrac{{{\left( 2{{F}_{0}} \right)}^{\dfrac{3}{2}}}}{{{F}_{0}}^{\dfrac{3}{2}}}\]
From this we will get the relation between the powers,
\[{P}'=2\sqrt{2}{{P}_{0}}\]

So, the correct answer is “Option D”.

Note:
Power of a body is given as the rate at which the body does a work. The unit of power is given as Watt. The body with more power can do more work. The lesser the power, the body can give away only a low amount of energy.