A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60 by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is : [g=10 m/s2] [figure]
- 5 3 J
- 5 J
- 5 103 J
- 10 J
Answer
(2) 5 J
Solution
N=mg 60 =1 10 12=5 N
f= N=0.1 5=0.5 N ( friction acts down the incline, opposite to motion)
Wf=f s=0.5 10=5 J
Answer: (2)