Question

A 20kg block moves with a constant speed of $10m{{s}^{-1}}$ on a smooth horizontal surface. What is the work done on the box?
a) 196J
b) 1960J
c) 980J
d) zero

Answer 1

Hint: It is given in the question that the block moves on a frictionless surface with constant speed. The work done on an object can basically be defined as the increase in its kinetic energy or the action of the force results in the displacement of the body. Hence we will first verify whether the above mentioned conditions are satisfied and accordingly draw the conclusion and select the correct alternatives provided.

Formula used:
$K.E=\dfrac{1}{2}m{{v}^{2}}\text{Joule}$
\[Work\text{ }done=F.d\text{ Joule}\]

Complete step-by-step answer:
Let us say that the mass of the block m=20kg and the speed of be $v=10m{{s}^{-1}}$. Therefore the kinetic energy of the block is given by,
$\begin{align}
  & K.E=\dfrac{1}{2}m{{v}^{2}}\text{J} \\
 & \Rightarrow K.E=\dfrac{1}{2}\times 20\times {{(10)}^{2}} \\
 & \Rightarrow K.E=\dfrac{1}{2}\times 20\times {{(10)}^{2}}=1000J \\
\end{align}$
It is further given to us the block is moving on a frictionless surface with constant velocity. Since the velocity ‘v’ is constant we can conclude that the change in kinetic energy is zero. Since change in kinetic energy is equal to work done on the box, we can imply that the work done on the box is zero
Hence the correct answer of the above question is option d.

So, the correct answer is “Option d”.

Note: Work done on a body is also defined as the product of the action of force(F) on the body with the displacement of the body (d) along the direction of the force. Mathematically this can be represented as \[Work\text{ }done=F.d\text{ J}\] . Force on the body is defined as the product of mass times the acceleration. Acceleration is nothing but the rate of change of velocity with time. Since the velocity of the box is constant i.e. it does not change with time we can conclude that force on the body is zero. And therefore from the above expression of work done we can conclude that the work done on the box is zero.