0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
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/s2)
  1. zero
  2. 100J
  3. 200J
  4. 300J

A
zero
B
100J
C
200J
D
300J
Solution
Verified by Toppr

Given,

Work done =300J, mass=2kg,height=10m,g=10m/s2
we know that the work done =fd

300=(mgsinθ+f)×hsinθ

Work-done against the friction:

f.hsinθ=300mgsinθ×hsinθ300mgh=3002×10×10=100J

983668_358825_ans_c934d73d8e7741cebb09b3ea74ad7540.PNG

Was this answer helpful?
71
Similar Questions
Q1
300 J of work done in sliding a 2 kg block up the inclined plane of height 10 m with constant velocity. The work done against friction is (Take g=10 m/s2)
View Solution
Q2
300J of work done in sliding a 2kg block up an inclined plane of height 10m. Taking g=10m/s2, work done against friction is:
View Solution
Q3
if 250J of work is done in sliding a 5kg block up an inclined plane of height 4m work done against fiction g=10ms2)
View Solution
Q4
300 J of work is done in sliding a 2 kg block up the inclined plane of height 10 m with constant velocity. The work done against friction is (Take g=10 m/s2)
View Solution