Question

A body of mass $1kg$ begins to move under the action of a time dependent force $\left( {2t\overrightarrow i + 3{t^2}\overrightarrow j } \right)N$ where $\hat i$ and $\hat j$ are unit vector along x and y axis. What power will be developed by the force at the time $t$?
A. $(2{t^3} + 3{t^4})W$
B. $(2{t^3} + 3{t^5})W$
C. $(2{t^{^2}} + 3{t^3})W$
D. $(2{t^2} + 4{t^4})W$

Answer 1

In simple words, Force is defined as the interaction between two objects while power is energy consumed over time (work). In the above question we want to calculate the power developed which we can calculate by multiplying force and velocity and also the time dependent force is given.

Complete step by step answer:
Firstly, finding acceleration by Newton’s law i.e., $F = ma$
Where $m$ is mass of the body and $a$is the acceleration
$\overrightarrow a = \frac{{\overrightarrow F }}{m}$
(because $m = 1kg$given)
Acceleration is defined as the rate of change of velocity i.e.,
$\overrightarrow a = \frac{{d\overrightarrow v }}{{dt}}$
Finding velocity by integrating,
$\overrightarrow v = \int_0^t {\overrightarrow a dt}$
Now,
$\overrightarrow v = \int_0^t {\left( {2t\overrightarrow i + 3{t^2}\overrightarrow j } \right)} dt$
$\overrightarrow v = {t^2}\overrightarrow i + {t^3}\overrightarrow j$
Now, using the formula of power,
$\overrightarrow P = \overrightarrow F \bullet \overrightarrow v$
Using the above calculated values and dot formula;
$\therefore\overrightarrow P = \left( {2{t^3} + 3{t^5}} \right)W$

Hence, the correct answer is B.

Note: The dot product is used to find the length of the vectors or to find the angle between the two vectors. While cross product is used to find a vector perpendicular to the plane spanned by two vectors. In the above question we are using dot product to find the power developed by the object.The dot product is based on the projection of one vector onto another. The dot product of two vectors $\overrightarrow a$ and $\overrightarrow b$ is given by where $\theta$ is the angle between the two vectors $\overrightarrow a$ and $\overrightarrow b$.