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Question: What will be the work done by a force \[\vec F = ( - 6{x^3}i)\,N\] is displacing a particle from \[x...

What will be the work done by a force F⃗=(−6x3i) N\vec F = ( - 6{x^3}i)\,N is displacing a particle from x=4 mx = 4\,m to x=−2 mx = - 2\,m ?
A. −240 J - 240\,J
B. 360 J360\,J
C. 120 J120\,J
D. 420 J420\,J

Explanation

Solution

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=∫F⃗(s)⋅dS⃗W = \int {\vec F(s)} \cdot d\vec S ,where F⃗\vec Fis the net force acting on the body which is dependant of the displacement, dS⃗d\vec S is the infinitesimal displacement of the body.

Complete step by step answer:
We have a force F⃗=(−6x3i)N\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=4mx = 4m to x=−2mx = - 2m. Hence, the particle moves along the negative X-axis. So, applied force is along the displacement of the body. Now, let for a dx⃗d\vec x displacement of the body work done is F⃗(x)⋅dx⃗\vec F(x) \cdot d\vec x. So, total work done will be, W=∫F⃗(x)⋅dx⃗W = \int {\vec F(x)} \cdot d\vec x

We can write, dx⃗=dxi^d\vec x = dx\hat i and F⃗=(−6x3i^)\vec F = ( - 6{x^3}\hat i) with limits from x=4x = 4 to x=−2x = - 2. So, the work done by the force will be, ∫x=4x=−2F⃗⋅dx⃗\int\limits_{x = 4}^{x = - 2} {\vec F} \cdot d\vec x
Putting the values we have, ∫x=4x=−2(−6x3i^)⋅(dxi^)\int\limits_{x = 4}^{x = - 2} {( - 6{x^3}\hat i)} \cdot (dx\hat i)
Performing the dot product we have,
W=−∫x=4x=−2(6x3)(dx)W = - \int\limits_{x = 4}^{x = - 2} {(6{x^3})} (dx)
Integrating we get,
W=−6x44∣4−2W = - 6\left. {\dfrac{{{x^4}}}{4}} \right|_4^{ - 2}
Putting the limit we get,
W=−6[(−2)44−444]W= - 6\left[ {\dfrac{{{{( - 2)}^4}}}{4} - \dfrac{{{4^4}}}{4}} \right]
Simplifying we get,
W=−6[164−2564]W = - 6\left[ {\dfrac{{16}}{4} - \dfrac{{256}}{4}} \right]
⇒W=−6[4−64]\Rightarrow W = - 6\left[ {4 - 64} \right]
⇒W=−6×−60\Rightarrow W = - 6 \times - 60
That will be equal to,
∴W=360\therefore W = 360
So, work done by the force will be 360 J360\,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 mS = 4 - ( - 2) = 6\,m