M = 1.8 kg

μ = 0.25

F acts at 45⁰ with horizontal.

s = displacement horizontally = 2 meters.

Frictional force along the horizontal in the direction opposite to the displacement

= Ff = μ * N = μ m g = 0.25 * 1.8 kg * 10 m/s² = 4.5 Newtons

Work done by the Frictional force = Wf = Ff . s = 4.5 * (-2 meters) = - 9 Joules

(it is negative as displacement is in the opposite direction to friction).

The component of force F, along the horizontal in the direction of s :

F Sin 45⁰ = F/√2

Since, the block moves with a uniform speed, there is no net force on the block. Hence

F/√2 = Ff = 4.5 Newtons

F = 4.5 √2 Newtons

Work done by F on the block = **F . s =** F/√2 * 2 = 9 Joules.

This is positive as the force does positive work on the block.

Gravitational force, ie., the weight and the normal force N are perpendicular to the displacement vector. Hence, the dot product is 0, for the work done.