Question

A position dependent force F = $4{x^3} + 3{x^2} + 2x - 4$Newton acts on a small body of 2 kg and displaces it from x = 0 to x = 2 m. Calculate the work done.

Answer 1

Since the force is position dependent, we cannot use the direct formula for work done which is, W = F.s. Instead, we will have to integrate the force from initial displacement to final displacement.

Complete step-by-step solution
Work done is a scalar quantity and is found by the dot product of force with displacement.
W = F.s
But in this question, the force acting on the body is dependent on position, so it changes at every point. Hence, the simple formula of dot product cannot be applied on the entire range of displacement.
However, if we consider very small displacement of the small body dx, we can assume that the force is constant for that very small displacement dx. So we can find out the small amount of work done dW by taking the dot product of the force acting on the body for the small displacement and the small displacement dx.
dW = F.dx
This small amount of work done can be added for each infinitesimal displacement for the entire displacement of the body from x = 0 to x = 2 m. This summation of small work done can be obtained by integrating it from x = 0 to x = 2 m.
$\int {dW = \int\limits_0^2 {F.dx} }$ …equation (1)
Substituting the value of force in terms of x in equation (1), we obtain,
$\int {dW = \int\limits_0^2 {(4{x^3} + 3{x^2} + 2x - 4)dx} }$
The expression on the right side of the equation can be separated and the integration of the individual terms can be added together to obtain the final expression. Separating the terms and integrating them individually, we get,
W = $\int {dW = \int\limits_0^2 {4{x^3}dx + \int\limits_0^2 {3{x^2}dx + \int\limits_0^2 {2xdx + \int\limits_0^2 {4dx} } } } }$ …equation (2)
The constant can be taken out of the integral. Using the formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ in equation (2), we obtain,
W = $4\left[ {\dfrac{{{x^4}}}{4}} \right]_0^2 + 3\left[ {\dfrac{{{x^3}}}{3}} \right]_0^2 + 2\left[ {\dfrac{{{x^2}}}{2}} \right]_0^2 + 4\left[ x \right]_0^2$
We do not add constant of integration while integrating in this case because it is a case of definite integration. We have to apply the limits of integration in the above expression.
W = $4\left( {\dfrac{{{2^4} - 0}}{4}} \right) + 3\left( {\dfrac{{{2^3} - 0}}{3}} \right) + 2\left( {\dfrac{{{2^2} - 0}}{2}} \right) + 4\left( {2 - 0} \right)$
On simplifying and solving, we obtain,
W = 16 + 8 + 4 + 8 N = 36 N

Hence, we obtain that the work done is 36 N.

Note: The mass of the body is an extra data and is not required to solve this question. Work is path independent. So, if the initial and final displacement remain the same, the work done will also remain the same, even if the path followed is different.