500 J of work is done in sliding a 2.5 kg block up an inclined plane of height 10 m. Taking g =10 m/s2, work done against friction is :
Answer & explanation
Correct answer: option 2
Work-Energy theorem : W = \(\Delta\) KE
\(\Delta\) KE = 0
500 – Wgravity – Wfriction = 0
Wgravity = mgh = 2.5m*10m/s2*10m = 250 J
⇒ Wfriction = 250 J