The gravitational field in a region is given by
$\to E = (2 i + 3 j) N / k g$
Find the work done by the gravitational field when a particle of mass $1 k g$ is moved on the line
$3 y + 2 x = 5$ from $(1 m, 1 m)$ to $(- 2 m, 3 m)$ .

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Solution

The force on the particle will be

$F = m E = (2 i + 3 j) N$

Direction of force is along line of slope $m_{1} = \frac{3}{2}$

Now, the direction of motion is along line of slope $m_{2} = - \frac{2}{3}$ .

Since $m_{1} m_{2} = - 1$ , hence force and displacement are perpendicular ro each other and thus the work done is 0.

Answer is 0J.

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The gravitational field in a region is given by→E=(2i+3j)N/kgFind the work done by the gravitational field when a particle of mass 1kg is moved on the line3y+2x=5 from (1m,1m) to (−2m,3m).

The force on the particle will beF=mE=(2i+3j)NDirection of force is along line of slope m1=32Now, the direction of motion is along line of slope m2=−23.Since m1m2=−1, hence force and displacement are perpendicular ro each other and thus the work done is 0.Answer is 0J.

The gravitational field in a region is given by
$\to E = (2 i + 3 j) N / k g$
Find the work done by the gravitational field when a particle of mass $1 k g$ is moved on the line
$3 y + 2 x = 5$ from $(1 m, 1 m)$ to $(- 2 m, 3 m)$ .