Answer
(C) $2$
Work Step by Step
Force is equal to $F=ma$. To find acceleration, use a kinematics equation relating acceleration, distance, initial velocity, and final velocity, which is $$v_f^2=v_o^2+2a\Delta x$$ Solving for $a$ yields $$a=\frac{v_f^2-v_o^2}{2\Delta x}$$ Since the object ends at rest, $v_f=0.00m/s$. $$a=\frac{-v_o^2}{2\Delta x}$$ This means that the original force is equal to $$F=\frac{mv_o^2}{2\Delta x}$$ If the new velocity $v_o'=2v_o$ and $\Delta x'=2\Delta x$, the new force $F'$ equals $$F'=\frac{m(2v_o)^2}{2(2\Delta x)}=\frac{4mv_o^2}{4\Delta x}=\frac{mv_o^2}{\Delta x}$$ $$=2(\frac{mv_o^2}{2\Delta x})=2F$$ Therefore, the force is doubled. This corresponds to choice $C$.
