300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s², work do

Question

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s², work done against friction is:

Accepted Answer

To find the work done against friction, we use the principle of conservation of energy. The total work done ($W_{total}$) by an external force to move the block up the incline is equal to the increase in the block's gravitational potential energy ($ΔU$) plus the work done against dissipative forces like friction ($W_{friction}$).

1.  **Calculate the increase in Potential Energy (ΔU):** The potential energy gained by raising a mass $m$ to a height $h$ is given by the formula $U = mgh$ , .
    $$ΔU = 2 	ext{ kg} 	imes 10 	ext{ m/s}^2 	imes 10 	ext{ m} = 200 	ext{ J}$$

2.  **Calculate work done against friction ($W_{friction}$):** Using the energy balance equation:
    $$W_{total} = ΔU + W_{friction}$$
    $$300 	ext{ J} = 200 	ext{ J} + W_{friction}$$
    $$W_{friction} = 300 	ext{ J} - 200 	ext{ J} = 100 	ext{ J}$$

Thus, 100 J of the total work was used to overcome friction.

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