Question

A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is              Joules (round off to the nearest integer).

Hint:

For force and displacement problems, break down the work integral into parts based on the components of the force. Always check for boundaries and evaluate integrals carefully.

Answer 1

The work done by the force is the line integral: \[ W = \int_0^4 x^2 (10 - x) dx + \int_0^2 y^2 dy \] Breaking down the integral: \[ W = \int_0^4 \left( x^2 (10 - x) \right) dx + \int_0^2 y^2 dy \] Now, solving the integrals: \[ \int_0^4 10x^2 - x^3 dx = \left[ \frac{10x^3}{3} - \frac{x^4}{4} \right]_0^4 = \frac{640}{3} - 64 = \frac{640}{3} - \frac{192}{3} = \frac{448}{3} \] \[ \int_0^2 y^2 dy = \left[ \frac{y^3}{3} \right]_0^2 = \frac{8}{3} \] Thus, the total work is: \[ W = \frac{448}{3} + \frac{8}{3} = \frac{456}{3} = 152 \, \text{Joules}. \]