Question

A block of mass 10kg is moving in x-direction with constant speed of 10m/s. It is subjected to a retarding force F=0.1x joule/meter during its travel from x=20cm to x-=30cm, Its final K.E will be
a)475J
b)450J
c)275J
d)250J

Answer 1

Hint: If we consider the above force of retardation, it is a variable force. Hence we have to integrate the work done by the force in order to stop the block during its travel from x = 20cm to x = 30cm. Since this work done will be equal to the reduction in the kinetic energy of the block, by taking the difference between initial kinetic energy and work done we will obtain the final kinetic energy.
Formula used:
$K.E=\dfrac{1}{2}m{{v}^{2}}J$
$dW=\int{Fdx}$

Complete answer:
Let us say a body of mass ‘m’ has some velocity ‘v’, Therefore its kinetic energy (K.E) is given by,
$K.E=\dfrac{1}{2}m{{v}^{2}}J$
In the above question it is given to us that the block moves with a velocity of 10m/s. Hence the kinetic energy of the block with mass 10kg equal to,
$\begin{align}
  & K.E=\dfrac{1}{2}m{{v}^{2}} \\
 & \Rightarrow K.E=\dfrac{1}{2}10kg\times {{(10m/s)}^{2}} \\
 & \Rightarrow K.E=500J \\
\end{align}$
Further it is given to us that the block is subjected to a retarding force. Since the force F is a variable force, The work small amount of work done in moving the block to a small distance (dx) is given by,
$dW=\int{Fdx}$
Force (F=0.1x) and it is acted on the block from x=20cm to x-=30cm. Therefore integrating the work done as per the above limits we get,
$\begin{align}
  & \int{dW}=\int\limits_{20m}^{30m}{0.1xdx} \\
 & \Rightarrow W=0.1\int\limits_{20m}^{30m}{xdx} \\
 & \Rightarrow W=0.1\left[ \dfrac{{{x}^{2}}}{2} \right]_{20m}^{30m} \\
 & \Rightarrow W=0.05\left[ {{30}^{2}}-{{20}^{2}} \right]=0.05\left[ 900-400 \right] \\
 & \Rightarrow W=25J \\
\end{align}$
This work done reduces the kinetic energy of the block as the force acted is in opposite direction of motion of the block, therefore the final K.E is equal to,
$\begin{align}
  & K.E(Final)=K.E-W \\
 & \Rightarrow K.E(Final)=500-25 \\
 & \Rightarrow K.E(Final)=475J \\
\end{align}$

Hence the correct answer of the above question is option a.

Note:
We said that the force subjected to the block as a variable force, as the force is a function of x. That means it changes with the displacement of the block. It is not necessary that the variable force is always a function of displacement. It may also vary with other parameters such as time, velocity etc.