Question

A 5 kg mass is accelerated from rest to 60 \[m{{s}^{-1}}\] in 1s. What force acts on it
\[\begin{align}
  & \text{A) 5}\times 60N \\
 & \text{B) (}\dfrac{\text{5}}{60}\text{)}\times 981N \\
 & \text{C) 6}{{\text{0}}^{2}}\times 52N \\
 & \text{D) (}\dfrac{\text{5}}{2}\text{)}\times {{60}^{2}}\times 981N \\
\end{align}\]

Answer 1

Hint: We need to understand the relation between mass of the body, the change in velocity of the body and the time taken for the change with the force acting on the body. We need to find the acceleration of the body using the given parameters.

Complete step by step solution:
We know from Newton's second law of motion that a body of mass ‘m’ can accelerate through a distance only when an external force acts upon the object. The force is the driving parameter of any body with mass. For a body to change its velocity from a given value to the final value, an external force is required, whereas for the body to continue in its velocity there should not be any force present affecting the body.
From the Newton’s second law, we get the equation for the force as the product of mass of the body and the acceleration it undergoes as –
\[F=ma\]
Where, F is the force on the body,
m is the mass of the body,
a is the acceleration of the body.
Now, we need to find the acceleration of the body from the given values of velocity and time. We are given that the body of mass 5 kg starts from an initial velocity of zero and ends up in a velocity of 60 \[m{{s}^{-1}}\] in a time interval of 1 s. So, we can find the acceleration as –
\[\begin{align}
  & a=\dfrac{v-u}{t} \\
 & \Rightarrow a=\dfrac{60-0}{1} \\
 & \therefore a=60m{{s}^{-2}} \\
\end{align}\]
Now, we can find the force acting on the mass as
\[\begin{align}
  & F=ma \\
 & \Rightarrow F=5kg\times 60m{{s}^{-2}} \\
 & \therefore F=5\times 60N \\
\end{align}\]
This is the required force for the body to attain the velocity.

The correct answer is option A.

Note:
We know that in practical cases, the external force applied to accelerate a body should be such that it is above the opposing forces acting on the body. The opposing forces can be frictional forces, viscous forces or air resistance depending on the situation.