Question

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

• A. 200 J
• B. 100 J
• C. zero
• D. 1000 J

Answer 1

Answer: (B)

100 J

Answer 2

Net work done in sliding a body up to a height $h$ on inclined plane
$=$ Work done against gravitational force $+$ Work done against frictional force
$\Rightarrow W=W_{g}+W_{f} \ldots$(i)
but $W=300\, J$
$W_{g}= m g h=2 \times 10 \times 10=200\, J$
Putting in E (i), we get
$300=200+W_{f}$
$\Rightarrow W_{f}=300-200=100\, J$