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

Work Calculation: Step-by-Step

1. Integral Formulation

Total work is computed as the sum of $x$ and $y$ line integrals: \[ W=\int_0^4 x^2(10-x)\,dx \;+\; \int_0^2 y^2\,dy \]

2. $x$-Integral Evaluation

\[ \int_0^4 (10x^2 - x^3)\,dx =\left[\frac{10x^3}{3}-\frac{x^4}{4}\right]_0^4 =\frac{640}{3}-64=\frac{448}{3} \]

3. $y$-Integral Evaluation

\[ \int_0^2 y^2\,dy =\left[\frac{y^3}{3}\right]_0^2 =\frac{8}{3} \]

4. Result Consolidation

\[ W=\frac{448}{3}+\frac{8}{3}=\frac{456}{3}=152 \]

Final Result:

$\boxed{152\ \text{J}}$