Question

What will be the work done by a force \[\vec F = ( - 6{x^3}i)\,N\] is displacing a particle from \[x = 4\,m\] to \[x = - 2\,m\] ?
A. \[ - 240\,J\]
B. \[360\,J\]
C. \[120\,J\]
D. \[420\,J\]

Answer 1

Hint: Work done on a body is given by the dot product of the force and displacement of the body. The formula of work done on a body is given by, \[W = \int {\vec F(s)} \cdot d\vec S\] ,where \[\vec F\]is the net force acting on the body which is dependant of the displacement, \[d\vec S\] is the infinitesimal displacement of the body.

Complete step by step answer:
We have a force \[\vec F = ( - 6{x^3}i)N\] which is acting on the body. Now, the displacement of the body is along the X-axis. We have given here that the particle moves from \[x = 4m\] to \[x = - 2m\]. Hence, the particle moves along the negative X-axis. So, applied force is along the displacement of the body. Now, let for a \[d\vec x\] displacement of the body work done is \[\vec F(x) \cdot d\vec x\]. So, total work done will be, \[W = \int {\vec F(x)} \cdot d\vec x\]

We can write, \[d\vec x = dx\hat i\] and \[\vec F = ( - 6{x^3}\hat i)\] with limits from \[x = 4\] to \[x = - 2\]. So, the work done by the force will be, \[\int\limits_{x = 4}^{x = - 2} {\vec F} \cdot d\vec x\]
Putting the values we have, \[\int\limits_{x = 4}^{x = - 2} {( - 6{x^3}\hat i)} \cdot (dx\hat i)\]
Performing the dot product we have,
\[W = - \int\limits_{x = 4}^{x = - 2} {(6{x^3})} (dx)\]
Integrating we get,
\[W = - 6\left. {\dfrac{{{x^4}}}{4}} \right|_4^{ - 2}\]
Putting the limit we get,
\[ W= - 6\left[ {\dfrac{{{{( - 2)}^4}}}{4} - \dfrac{{{4^4}}}{4}} \right]\]
Simplifying we get,
\[W = - 6\left[ {\dfrac{{16}}{4} - \dfrac{{256}}{4}} \right]\]
\[\Rightarrow W = - 6\left[ {4 - 64} \right]\]
\[\Rightarrow W = - 6 \times - 60\]
That will be equal to,
\[\therefore W = 360\]
So, work done by the force will be \[360\,J\] .

Hence, option B is correct.

Note: If the force acting up on the body was a constant force then, no integration need not be performed to find the work done by the force. Since, then the force will be a constant of and the integration will be on the displacement only, which is same as the net displacement of the particle that is \[S = 4 - ( - 2) = 6\,m\]