A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter./find the work done by this force during a displacement form x=0 tox=2.0m

Options

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

Answer

C

Explanation

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