The force on a particle as the function of displacement x(in x-direction) is given by $$F=10+0.5x$$
The work done corresponding to displacement of particle from $$x=0$$ to $$x=2$$ unit is?
Correct option is C. $$21$$ J
Given that $$F=10+0.5x$$
Work done corresponding to the particle from $$x=0$$ to $$x=2$$ unit is, $$W= \int_{0}^{2} F.dx= \int_{0}^{2} (10+0.5x)dx= [10x+ \dfrac{0.5x^2}{2}]_0^2= 20+1= 21 J$$