Question

A body is displaced from (0, 0) to (1m, 1m) along the path x = y by a force $\vec{F} = (x^2\hat{i}+y\hat{j})$ N. The work done by this force will be PJ. Find the value of P

Answer 1

5/6

Answer 2

The work done by a variable force $\vec{F}$ over a displacement $d\vec{r}$ is given by the line integral:

$W = \int \vec{F} \cdot d\vec{r}$

Given force: $\vec{F} = (x^2\hat{i}+y\hat{j})$ N

Differential displacement: $d\vec{r} = dx\hat{i} + dy\hat{j}$ m

First, calculate the dot product $\vec{F} \cdot d\vec{r}$:

$\vec{F} \cdot d\vec{r} = (x^2\hat{i}+y\hat{j}) \cdot (dx\hat{i} + dy\hat{j})$

$\vec{F} \cdot d\vec{r} = x^2 dx + y dy$

The body is displaced from (0, 0) to (1m, 1m) along the path x = y.

Since x = y, it implies that $dx = dy$.

Substitute y = x and dy = dx into the expression for work done:

$W = \int_{(0,0)}^{(1,1)} (x^2 dx + x dx)$

$W = \int_{(0,0)}^{(1,1)} (x^2 + x) dx$

The limits of integration for x will be from 0 to 1.

$W = \int_{0}^{1} (x^2 + x) dx$

Now, perform the integration:

$W = \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_{0}^{1}$

Evaluate the definite integral using the limits:

$W = \left( \frac{1^3}{3} + \frac{1^2}{2} \right) - \left( \frac{0^3}{3} + \frac{0^2}{2} \right)$

$W = \left( \frac{1}{3} + \frac{1}{2} \right) - (0)$

To add the fractions, find a common denominator, which is 6:

$W = \frac{2}{6} + \frac{3}{6}$

$W = \frac{5}{6} \text{ J}$

The work done is given as PJ.

Comparing $W = \frac{5}{6}$ J with $W = PJ$, we find:

$P = \frac{5}{6}$