Question

A body of mass 1 kg begins to move under the action of time dependent force $\vec F = (2t\hat i + 3{t^2}\hat j)N$ Where $\hat i$ and $\hat j$ are unit vectors along $x$ and $y$ axes. What power will be developed by the force at the time, t?
(A) $(2{t^2} + 3{t^2})W$
(B) $(2{t^4} + 4{t^4})W$
(C) $(2{t^3} + 3{t^4})W$
(D) $(2{t^3} + 3{t^5})W$

Answer 1

Hint: By using the Newton’s second law of motion for finding the acceleration and by using the relation between the force and velocity and also using the relation between the power, work and time, then the power developed by the force at the time is determined.

Formula used:
Newton’s second law,$\vec F = m\vec a$
Where $\vec F$ is the force, $m$ is the mass of the body and $\vec a$ is the acceleration of the body
Power,$P = \dfrac{W}{t}$
Where, W is the total work done and t is the total time taken
By substituting for work, $W = Fx\cos \theta $
Where $x$ is the displacement
Power will become,
$P = \dfrac{{Fx\cos \theta }}{t}$
We know that
$v = \dfrac{x}{t}$ Where $v$ is the velocity
Substituting this value in the equation of power we will get
$ P = \vec Fv\cos \theta = \vec F \cdot \vec v $

Complete step by step solution:
Given that,
The time dependent force $\vec F = (2t\hat i + 3{t^2}\hat j)N$
Mass $M = 1kg\;$
First, we have to find the acceleration by using newton’s second law of motion
$\Rightarrow \vec F = M\vec a$
By rearranging this equation, we will get the acceleration as,
$\Rightarrow \vec a = \dfrac{{\vec F}}{M}$
Substituting the value of force and mass,
$\Rightarrow \vec a = \dfrac{{(2t\hat i + 3{t^2}\hat j)}}{{1kg}}$
$\Rightarrow \vec a = (2t\hat i + 3{t^2}\hat j)$
As we know that the acceleration is the rate of change of velocity by using this,
$\Rightarrow \vec a = \dfrac{{d\vec v}}{{dt}}$
By rearranging this equation,
$\Rightarrow d\vec v = \vec adt$
Taking integral on both sides,
\[\Rightarrow \int {d\vec v = \int {\vec adt} } \]
Substituting the value of acceleration and integrating we will get,
\[\Rightarrow \vec v = \int\limits_0^t {(2t\hat i + 3{t^2}\hat j)dt} \]
By applying the sum rule of integration,
\[\Rightarrow \vec v = \int\limits_0^t {2t\hat idt + \int\limits_0^t {3{t^2}\hat jdt} } \]
\[\Rightarrow \vec v = {t^2}\hat i + {t^3}\hat j\]
Now use the equation of power,
$\Rightarrow P = \vec F \cdot \vec v$
Substitute the values of force and velocity to find the power,
$\Rightarrow P = (2t\hat i + 3{t^2}\hat j) \cdot ({t^2}\hat i + {t^3}\hat j)$
Take the dot product of the above equation,
Now we will get
$\Rightarrow P = (2{t^3} + 3{t^5})$

Thus, the option (D) is correct.

Note: Use newton’s second law to find the acceleration and arrange the equation of power to get the relation between power, velocity and the force. Use the sum rule of integration for the simplification. While taking dot product the coefficient of the $\hat i$ is multiplied with the coefficient of $\hat i$ and it is similar to the $\hat j$ also.