A force `F=(10+0.5 x)` acts on a particle in the x-direction. What would be the work done by this force during a displacement from `x=0` to `x=2m` (F is in newton and x in metre)

Options

A. 31.5 J
B. 63 J
C. 21 J
D. 42 J

Answer

C

Explanation

Given, force, `F=10+0.5 x=10 +1/2 x`
Let during small displacement the work done by the force is `dW=Fdx`.
So, work done during displacement from `x=0` to `x=2` is
`W=int_(0)^(w) d W=int_(0)^(2) F dx=int_(0)^(2) (10 +1/2x) dx`
`=10 [x]_(0)^(2)+[x^(2)/4]_(0)^(2)=21 J`