Question 234: Work, Energy and Power & Circular Motion

Question

A force $(3 x^2+2 x-5) \mathrm{N}$ displaces a body from $x=2 \mathrm{~m}$ to $x=4 \mathrm{~m}$. Work done by this force is ________ J.

Accepted Answer

To find the work done by the force when the body is displaced from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$, we use the formula for work done by a variable force in one dimension, which is the integral of the force with respect to displacement:
$ W = \int_{x_1}^{x_2} F \, dx $
Given the force $F(x) = (3x^2 + 2x - 5) \, \mathrm{N}$ and the limits of integration from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$, we can substitute these values into the equation:
$ W = \int_{2}^{4} (3x^2 + 2x - 5) \, dx $
Calculating the integral, we get:
$ W = \left[\frac{3x^3}{3} + \frac{2x^2}{2} - 5x\right]_2^4 $
This simplifies to:
$ W = \left[x^3 + x^2 - 5x\right]_2^4 $
Substituting the upper limit ($x = 4$) and then the lower limit ($x = 2$) into the antiderivative, and subtracting the latter from the former, we get:
$ W = \left[(4)^3 + (4)^2 - 5(4)\right] - \left[(2)^3 + (2)^2 - 5(2)\right] $
$ W = (64 + 16 - 20) - (8 + 4 - 10) $
$ W = 60 - 2 $
$ W = 58 \, \mathrm{J} $
Therefore, the work done by the force as the body displaces from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$ is $58 \, \mathrm{J}$.

Work, Energy and Power & Circular Motion

A force (3 x^2+2 x-5) \mathrm{N} displaces a body from x=2 \mathrm{~m} to x=4 \mathrm{~m}. Work done by this force is ________ J.

Integer / Numeric Answer Type