Question

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

• A. zero
• B. 100 J
• C. 200 J
• D. 300 J

Answer 1

Answer: (B)

100 J

Answer 2

Work done against friction is equal to
$W' = W - U$

where $U$ is potential energy, $W$ the work done in sliding the block up the inclined plane.
$U = mgh$
$=2\times 10\times 10=200\,J$
$W = 300\, J$
$\therefore W' = 300 - 200 = 100\, J$