$300 J$ of work is done in sliding a $2 k g$ block up an inclined plane of height $10 m$ . The work done against friction is (Take $g = 10 m / s$ $^{2}$ )

Question

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Answer 1

$Step 1: Draw the free body diagram$

$Step 2: Write the equation for work done$

Let $W_{f}$ be the work done against friction

The total work done on a particle is equal to the change in its mechanical energy

$W = W_{f} + P E$

$= W_{f} + m g h . . . . . . . . (1)$

$Step 3: Work done against the friction$

From equation (1)

$W_{f} = W - m g h$

Substituting the values :

$W_{f} = 300 J - (2 k g \times 10 m / s^{2} \times 10 m) = 100 J$

Hence option $(B)$ is correct.

Answer 2

Step 1: Draw the free body diagram

Step 2: Write the equation for work done

Let

W_{f}

be the work done against friction

The total work done on a particle is equal to the change in its mechanical energy

W = W_{f} + P E

= W_{f} + m g h . . . . . . . . (1)

Step 3: Work done against the friction

From equation (1)

W_{f} = W - m g h

Substituting the values :

W_{f} = 300 J - (2 k g \times 10 m / s^{2} \times 10 m) = 100 J

Hence option

(B)

is correct.