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 done by the force — stepwise evaluation

1. Integral setup 

The total work is the line integral split into the $x$– and $y$–parts: \[ W=\int_0^4 x^2(10-x)\,dx \;+\; \int_0^2 y^2\,dy \]

2. Evaluate the $x$–integral

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

3. Evaluate the $y$–integral

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

4. Combine results

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

Final Answer:

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

Answer 2

To calculate the total work done by the force, we evaluate the line integral along the given path:

1. Integral Setup:
The work is expressed as the sum of two integrals:

\[ W = \int_0^4 x^2(10 - x)\,dx + \int_0^2 y^2\,dy \]

2. Solving the First Integral (x-component):
\[ \int_0^4 (10x^2 - x^3)\,dx = \left[ \frac{10x^3}{3} - \frac{x^4}{4} \right]_0^4 \]
Evaluating at bounds:
\[ = \left(\frac{10(64)}{3} - \frac{256}{4}\right) - 0 = \frac{640}{3} - 64 = \frac{448}{3} \]

3. Solving the Second Integral (y-component):
\[ \int_0^2 y^2\,dy = \left[ \frac{y^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3} \]

4. Combining Results:
\[ W = \frac{448}{3} + \frac{8}{3} = \frac{456}{3} = 152 \]

Physical Interpretation:
- The first integral represents work done along the x-axis (0 to 4) with varying force
- The second integral represents work done along the y-axis (0 to 2)
- Units are consistent in Joules for work

Final Answer:
The total work done is \(152 \text{ Joules}\).