Consider the adjacent diagram. As the block M is at rest.

Hence, f = frictional force = $$Mg\sin\theta$$

The force of friction acting between the block and incline opposes the tendency of sliding of the block. Since, block is not in motion, therefore, no work is done by the force of friction. Hence, no dissipation of energy takes place.

When the elevator is descending, then electric power is required to prevent it from falling freely under gravity.

Also, as the weight inside the elevator increases, its speed of descending increases, therefore, there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with large velocity.

(a) Force is applied on the body to lift it in upward direction and displacement of the body is also in upward direction, therefore, angle between the applied force and displacement is $$\theta=0^{\circ}$$

$$\therefore$$ Work done by the applied force

$$W=F s \cos \theta=F s \cos 0^{\circ}=F s \left(\because \cos 0^{\circ}=1\right)$$

i.e., $$\qquad W=$$ Positive

(b) The gravitational force acts in downward direction and displacement in upward direction, therefore, angle between them is $$\theta=180^{\circ}$$.

$$\therefore$$ Work done by the gravitational force

$$W=F s \cos 180^{\circ}=-F s \quad\left(\because \cos 180^{\circ}=1\right)$$

Force of gravity acts on the car vertically downward while car is moving along horizontal road, i.e., angle between them is $$90^{\circ}$$.

Work done by the car against gravity

$$W=F s \cos 90^{\circ}=0 \quad\left(\because \cos 90^{\circ}=0\right)$$

No, total mechanical energy of the body falling freely under gravity is not conserved, because a small part of its energy is utilised against resistive force of air, which is non-conservative force. In this condition, gain in $$\mathrm{KE}<$$ loss in PE.