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

Correct Answer: 58

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: \[ W = 60 - 2 = 58 \, \mathrm{J}. \]