Question

A force $(3x^2 + 2x - 5) \, \text{N}$ displaces a body from $x = 2 \, \text{m}$ to $x = 4 \, \text{m}$. The work done by this force is _________ J.

Answer 1

Work done by a variable force is given by:
$W = \int_{x_1}^{x_2} F(x) \, dx.$
Here, $F(x) = 3x^2 + 2x - 5$, $x_1 = 2$, and $x_2 = 4$.
Substituting:
$W = \int_{2}^{4} (3x^2 + 2x - 5) \, dx.$

Evaluate the integral:
$\int (3x^2 + 2x - 5) \, dx = x^3 + x^2 - 5x.$

Substitute the limits: $W = \left[ x^3 + x^2 - 5x \right]_2^4.$
At $x = 4$: $W_4 = 4^3 + 4^2 - 5(4) = 64 + 16 - 20 = 60.$ At $x = 2$:
$W_2 = 2^3 + 2^2 - 5(2) = 8 + 4 - 10 = 2.$

The total work done: $W = 60 - 2 = 58 \, \mathrm{J}.$

Answer 2

Work done by a variable force is given by:
$W = \int_{x_1}^{x_2} F(x) \, dx.$
Here, $F(x) = 3x^2 + 2x - 5$, $x_1 = 2$, and $x_2 = 4$.
Substituting:
$W = \int_{2}^{4} (3x^2 + 2x - 5) \, dx.$

Evaluate the integral:
$\int (3x^2 + 2x - 5) \, dx = x^3 + x^2 - 5x.$

Substitute the limits: $W = \left[ x^3 + x^2 - 5x \right]_2^4.$
At $x = 4$: $W_4 = 4^3 + 4^2 - 5(4) = 64 + 16 - 20 = 60.$ At $x = 2$:
$W_2 = 2^3 + 2^2 - 5(2) = 8 + 4 - 10 = 2.$

The total work done: $W = 60 - 2 = 58 \, \mathrm{J}.$