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 = 10m/s${}^2$)

Answer 1

Given,

Work done $=300J$, $mass=2kg,height=10m,g=10m/s^2$

we know that the work done $=fd$

$300=(mg\sin \theta +f)\times \frac{h}{\sin \theta }$

Work-done against the friction:

$f.\frac{h}{\sin \theta }=300-mg\sin \theta \times \frac{h}{\sin \theta }\Rightarrow 300-mgh=300-2\times 10\times 10=100J$