Question

The gravitational field in a region is given by E=(2→i+3→j) N kg-1. Show that no work is done by the gravitational field when a particle is moved on the line 3y + 2x = 5.
[Hint If a line y = mx + c makes angle θ with the X-axis, m = tan θ.]

Answer 1

No work will be done when the particle will be on the line.

E = 2i + 3j (Given)

Equation of line = 3y + 2x = 5 (Given)

The field will be represented as -  

Tan Ф = 3/2

The line 3y + 2x= 5 will be represented as -  

Tan Ф = - 2/3

m1m2 = -1

Since, the direction of the field and the displacement are both perpendicular, no work will be done  by the particle that is on line.

Answer 2

Explanation:

• It is given that the gravitational field in the region is $E=(2 \hat{t}+3 \hat{y}) \mathrm{N} k g^{-1}$ and the particle is moved on the line $3 y+2 x=5$.  
• The slope of the electric field can be written as, $m_{1}=\tan \theta_{1}, \text { which implies } m_{1}=\frac{3}{2}$, The slope of the line can be written as $m_{2}=\tan \theta_{2}=-\frac{2}{3}$                  
• Hence, $m_{1} m_{2}=-1$.  As the direction of the gravitational field and the displacement of the particle are perpendicular to each other, no work can be done by the gravitational field when the particle is moved on the line $3 y+2 x=5$.