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 to x=2.0m

Explanation

Here, `F=(10+0.50x)`
Small amount of word done in moving the particle through a samll distance dx is
`dW=hat(F).d hatx=(10+0.5x)dx`
Total work done, `W=int_(x=0)^(x=2)(10+00.5x)dx`
`W=[10x +0.5(x^(2))/(2)]_(0)^(2)`
`=10(2-0)+(0.5)/(2)(2^(2)-0)=20+1`
`=21joule`