The Gravitational Field In a region is given by E = (3i-4j) N/kg Find the work done (in joule) in Displacing a particle by 1m along the line 4y= 3x+9.

To solve this problem, we need to use the formula for work done, which is the dot product of the force and the displacement vectors.

1. First, we need to find the force acting on the particle. The force is given by the product of the gravitational field and the mass of the particle. However, since the mass of the particle is not given, we will assume it to be 1 kg for simplicity. Therefore, the force vector F is the same as the gravitational field E, which is (3i - 4j) N.

2. Next, we need to find the displacement vector. The line along which the particle is displaced is given by 4y = 3x + 9. We can rewrite this as y = (3/4)x + 9/4. The slope of this line is 3/4, which means that for every 4 units the particle moves in the x direction, it moves 3 units in the y direction. Therefore, if the particle is displaced by 1 m along this line, its displacement in the x direction is 4/5 m and in the y direction is 3/5 m. Therefore, the displacement vector D is (4/5)i + (3/5)j m.

3. Finally, we can find the work done by taking the dot product of the force and displacement vectors. The dot product of two vectors (a1i + b1j) and (a2i + b2j) is given by a1a2 + b1b2. Therefore, the work done W is (34/5 - 43/5) J = 0 J.

Therefore, the work done in displacing the particle by 1 m along the given line is 0 Joules.