Question

A body of mass $2Kg$ is driven by an engine delivering a constant power of $1J/s$.The body starts from rest and moves in a straight line. After $9$ seconds, the body has moved a distance (in m) ………………….?

Answer 1

Hint: Since the problem is based on kinetic energy and work-energy theorem, consider the effect of power used to drive the body on kinetic energy (i.e., velocity). As we all know that the parameters vary with each other hence, analyze every aspect of the solution needed for the question and then present the answer with a proper explanation.

Formula used:
The work-energy theorem is represented as $W = {K_f} - {K_i}$.
And, the formula of power is given by:
$\text{Power(P)} = \dfrac{\text{Work done(W)}}{\text{Time(t)}}$

Complete step by step solution:
In this question, we have given the mass of a body, $m = 2kg$
Body drove by a constant power of $P = 1J/s$
Applying the Work-Energy theorem, $W = {K_f} - {K_i}$
Here, Initial Kinetic Energy be ${K_i}$and final kinetic energy be ${K_f}$
But, according to the question as the body starts from rest, the initial kinetic energy ${K_i} = 0$
$\therefore W = {K_f} = \dfrac{1}{2}m{v^2}$ … (1)
Also, we know $\text{Power(P)} = \dfrac{\text{Work done(W)}}{\text{Time(t)}}$ , then:
$W = P \times t$ … (2)
Equating (1) and (2), we get
$\dfrac{1}{2}m{v^2} = P \times t \\$
$\Rightarrow {v^2} = \dfrac{{2 \times P \times t}}{m} \\$
$\Rightarrow v = \sqrt {\dfrac{{2Pt}}{m}} \\$
But, $v = \dfrac{{ds}}{{dt}}$ means the rate of change in distance. Substituting this value in the above expression we get
$\dfrac{{ds}}{{dt}} = \sqrt {\dfrac{{2Pt}}{m}} \\$
$\Rightarrow ds = \sqrt {\dfrac{{2Pt}}{m}} \cdot dt \\$
Integrating both sides by applying power rule $\left( {\therefore \int {{x^n}.dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right)$ , we obtain:
$\Rightarrow \int\limits_0^s {ds} = \int\limits_0^t {\sqrt {\dfrac{{2Pt}}{m}} \cdot dt} \\$
$\Rightarrow (s - 0) = \sqrt {\dfrac{{2P}}{m}} \left( {\dfrac{{2{{(t - 0)}^{\dfrac{3}{2}}}}}{3}} \right) \\$
$\Rightarrow s = \sqrt {\dfrac{{2P}}{m}} \left( {\dfrac{{2{t^{\dfrac{3}{2}}}}}{3}} \right) \\$
Substituting the values of $m$ and $P$ from question, we get
At $t = 9s$,
$\Rightarrow s = \sqrt {\dfrac{{2(1)}}{{(2)}}} \left( {\dfrac{{2{{(9)}^{\dfrac{3}{2}}}}}{3}} \right) \\$
$\Rightarrow s = \dfrac{{2{{(3)}^3}}}{3} \\$
Thus, the distance after $9$ seconds will be:
$ \therefore s = 18m$

Hence, after $9$ seconds, the body has moved a distance of $18m$.

Note: This is a numerical problem based on the application of the work-energy theorem hence, it is essential that the given question is to be analyzed very carefully to give an accurate solution. While solving the question, take power used as an important factor and correlate the terms used with each other that might help in the solution.