Question

A body of mass 10kg at rest is acted upon simultaneously by two forces 4N and 3N at right angles to each other. The kinetic energy of the body at the end of 10 sec is?
A. 100 J
B. 300 J
C. 50 J
D. 125 J

Answer 1

Hint:We start proceeding with this problem by understanding the term kinetic energy. The kinetic energy of a body is defined as the energy possessed by a body when it is in motion.

Formula Used:
In order to find the formula for kinetic energy we have,
\[K.E = \dfrac{1}{2}m{v^2}\]
Where, m is mass of a body and v is velocity of the body.

Complete step by step solution:
Consider a body of mass 10kg which is at rest acted upon simultaneously by two forces 4N and 3N at right angles to each other. Given \[m = 10kg\], \[\overrightarrow {{F_1}} = 4N\] and \[\overrightarrow {{F_2}} = 3N\]. We need to find the kinetic energy of the body at the end of \[t = 10\,s\]. Since the body is acted upon simultaneously by two forces 4N and 3N at right angles, the net force is given by,
\[F = \sqrt {{4^2} + {3^2}} \]
\[\Rightarrow F = \sqrt {25} \]
\[\Rightarrow F = 5\,N\]
We know that,
\[a = \dfrac{F}{m}\]
Substitute the value of force and mass then we get,
\[a = \dfrac{5}{{10}}\]
\[\Rightarrow a = 0.5\,m{s^{ - 2}}\]

Now, we have formula to find the kinetic energy and is given by,
\[K.E = \dfrac{1}{2}m{v^2}\]
Since they have not given the value of v, we know that \[v = at\]. Substitute this value in the above equation we get,
\[K.E = \dfrac{1}{2}m{\left( {at} \right)^2}\]
\[\Rightarrow K.E = \dfrac{1}{2} \times 10{\left( {0.5 \times 10} \right)^2}\]
\[\Rightarrow K.E = \dfrac{1}{2} \times 10\left( {25} \right)\]
\[\therefore K.E = 125\,J\]
Therefore, the kinetic energy of the body at the end of 10 sec is 125 J.

Hence, option D is the correct answer.

Note:We don’t need to worry even if the value of velocity is not given. The velocity can also be written as \[v = at\] and using this solve for the kinetic energy.