Question

A force of \( (6x^2 - 4x + 3) \, \text{N} \) acts on a body of mass 0.75 kg and displaces it from \( x = 5 \, \text{m} \) to \( x = 2 \, \text{m} \). The work done by the force is

• A. 201 J
• B. 215 J
• C. 229 J
• D. 307 J

Hint:

For variable forces, the work done is found by integrating the force function over the displacement. Ensure to evaluate the definite integral properly for the correct limits.

Answer 1

The Correct Option is A

Step 1: The work done by a variable force is given by the integral of the force over the displacement: \[ W = \int_{x_1}^{x_2} F(x) \, dx \] Substitute the given force \( F(x) = 6x^2 - 4x + 3 \) and limits \( x_1 = 5 \) and \( x_2 = 2 \): \[ W = \int_{5}^{2} (6x^2 - 4x + 3) \, dx \] Step 2: Now, solve the integral: \[ \int (6x^2 - 4x + 3) \, dx = 2x^3 - 2x^2 + 3x \] Evaluating this from \( x = 5 \) to \( x = 2 \): \[ W = \left[ 2(2)^3 - 2(2)^2 + 3(2) \right] - \left[ 2(5)^3 - 2(5)^2 + 3(5) \right] \] \[ W = \left[ 2(8) - 2(4) + 6 \right] - \left[ 2(125) - 2(25) + 15 \right] \] \[ W = \left[ 16 - 8 + 6 \right] - \left[ 250 - 50 + 15 \right] \] \[ W = 14 - 215 = 201 \, \text{J} \] Thus, the work done by the force is 201 J.