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/s2, work done against friction is :

• A. 50 J
• B. 100 J
• C. zero
• D. 150 J

Answer 1

Answer: (B)

100 J

Answer 2

Work Done Against Friction: Work done against friction can be calculated using the formula W = F$\times$d, where F is the force of friction and d is the displacement.
Given that 300 J of work is done to slide the block up an inclined plane, and taking g = 10 m/s2, the gravitational force component along the incline is mg = 2 kg$\times$10 m/s2 = 20 N.
Work done against gravity is Wgravity = m$\times$g $\times$ h = 2 kg $\times$10 m/s2$\times$10 m = 200 J.
Work done against friction is Wfriction = Total work - Work against gravity = 300 J - 200 J = 100 J.