Question

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10m .the work done against friction is:
(A) zero
(B) 100 J
(C) 300J
(D)200J

Answer 1

Answer:

(B) 100 J

Explanation:

Given,

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

2

we know that the work done =fd

300=(mgsinθ+f)×

sinθ

h

Work-done against the friction:

f.

sinθ

h

=300−mgsinθ×

sinθ

h

⇒300−mgh=300−2×10×10=100J