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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) 1001 (b) Zero (c) 1000 J (d) 200 J



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)    1001


(b)    Zero


        (c)    1000 J      


(d)    200 J


Grade:upto college level

2 Answers

Simran Bhatia
348 Points
7 years ago

(a)

Work done against gravity = mg sin q × d

= 2 × 10 × 10    (d sinq = 10)

= 200 J

Actual work done = 300 J

Work done against friction = 300 - 200 = 100 J

Muntazir
15 Points
2 years ago
Loss in potential energy =mgh =2×10×10=200j gain in kinetic enegy =work done=300j work against friction =300-200=100

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