Question

A block is constrained to move along the x-axis under a force of \[F = - 2x\], here F is in newton and x is in meter. Find the work done by this force when the block is displaced from \[x = 2\,m\] to \[x = - 4\,m\].
A. \[ - 4\,{\text{J}}\]
B. \[ - 8\,{\text{J}}\]
C. \[ - 12\,{\text{J}}\]
D. \[ - 16\,{\text{J}}\]

Answer 1

Hint: The given force has variable force nature. Use the formula for work done by the variable force and substitute the values of limits that given in the question. The sign of work done should be negative.

Formula used:
\[W = \int\limits_{{x_1}}^{{x_2}} {F\,dx} \]
Here, F is the variable force and dx is the small displacement.

Complete step by step solution:
We know the formula for work done by the variable force,
\[W = \int\limits_{{x_1}}^{{x_2}} {F\,dx} \]
Here, F is the variable force and dx is the small displacement, \[{x_1}\] and \[{x_2}\] are the initial and final positions of the body.
Substituting\[F = - 2x\], \[{x_1} = 2\,m\] and \[{x_2} = - 4\,m\] in the above equation, we get,
\[W = \int\limits_2^{ - 4} {\left( { - 2x} \right)\,dx} \]
\[ \Rightarrow W = - 2\left( {\dfrac{{{x^2}}}{2}} \right)_2^{ - 4}\]
\[ \Rightarrow W = - \left( {{{\left( { - 4} \right)}^2} - {{\left( 2 \right)}^2}} \right)\]
\[ \Rightarrow W = - \left( {16 - 4} \right)\]
\[ \Rightarrow W = - 12\,{\text{J}}\]
Therefore, the work done by the force to displace the body from \[x = 2\,m\] to \[x = - 4\,m\] is \[ - 12\,{\text{J}}\].

So, the correct answer is “Option C”.

Additional Information:
The work done by the spring force is can be calculated as follows,
\[W = \int {F\,dx} \]
\[ \Rightarrow W = \int\limits_{{x_i}}^{{x_f}} { - kx} \,dx\]
\[ \Rightarrow W = - k\left( {\dfrac{{{x^2}}}{2}} \right)_{{x_i}}^{{x_f}}\]
\[ \Rightarrow W = - k\dfrac{{\left( {x_f^2 - x_i^2} \right)}}{2}\]
We can also calculate the work done by this force using the above equation.

Note:
While solving these types of questions, integration is necessary. To solve this question, you must know the difference between variable force and constant force. If the constant force acts on the body, then you don’t need to integrate the force term.