Question 233: 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