`300 J`of work is done in slide a `2kg` block up an inclined plane of height `10m`. Take `g = 10 m//s^(2)`, work done against friction is

Options

A. `200 J`
B. `100 J`
C. zero
D. `1000 J`

Answer

B

Explanation

Net Net work done in sliding a body up to a height `h` on inclined plane
`= `Work done against gravitatioinal force
+ Work done against frictional force
` implies W = W_(g) + W_(f)` (i)
but `W = 300J`
`W_(g) + mgh = 2 xx 10 xx 10 = 200 J`
Putting in Eq (i) we get
`300 = 200 + W_(f)`
`W_(f) = 300 - 200 = 100J`