A force F =( 2i^−3j^+k^ ) N is acting on an object of mass 2kg which is moving from point A (0, 1m, 2m) to point B(3m, 0, 3m) then work done by the force on the object will be

The work done by a force on an object is given by the dot product of the force vector and the displacement vector.

Step 1: Find the displacement vector. The displacement vector is the final position minus the initial position. So, the displacement vector D = B - A = (3m, 0, 3m) - (0, 1m, 2m) = (3m, -1m, 1m).

Step 2: Calculate the dot product of the force and displacement vectors. The dot product of two vectors is the sum of the products of their corresponding components. So, F.D = (2N * 3m) + (-3N * -1m) + (1N * 1m) = 6J + 3J + 1J = 10J.

Therefore, the work done by the force on the object is 10 Joules.

The work done by a force on an object is given by the dot product of the force vector and the displacement vector.

Step 1: Calculate the displacement vector. The displacement vector is the final position minus the initial position. In this case, the initial position is point A (0, 1m, 2m) and the final position is point B (3m, 0, 3m). So, the displacement vector is (3-0)i^ + (0-1)j^ + (3-2)k^ = 3i^ - j^ + k^.

Step 2: Calculate the dot product of the force vector and the displacement vector. The force vector is F = 2i^ - 3j^ + k^ and the displacement vector is 3i^ - j^ + k^. The dot product of two vectors is the sum of the products of their corresponding components. So, the dot product of F and the displacement vector is (23) + (-3-1) + (1*1) = 6 + 3 + 1 = 10 Joules.

So, the work done by the force on the object is 10 Joules.