Answer
$34m$
Work Step by Step
All units must be in SI units. So, the speed must be converted to meters per second using dimensional analysis. $$\frac{48km}{1hr}\times \frac{1hr}{60min} \times \frac{1min}{60s} \times \frac{1000m}{1km}$$ $$=13m/s$$ The minimum force needed for the object to move under friction is equal to $$F_{min}=\mu _smg$$ Also note that force is defined as $$F=ma$$ Therefore, acceleration is equal to $$a=\frac{F}{m}$$ Substituting the formula for force yields an acceleration of $$a=\frac{\mu_smg}{m}=\mu_sg$$ Substituting known values of $\mu_s=0.25$ and $g=9.8m/s^2$ yields a maximum acceleration of $$a=(0.25)(9.8m/s^2)=2.5m/s^2$$ To find the minimum distance for stopping, use a kinematics equation using acceleration, distance, initial velocity, and final velocity. This is $$v_f^2=v_o^2+2a\Delta x$$ Solving for $\Delta x$ yields $$\Delta x=\frac{v_f^2-v_o^2}{2a}$$ Substituting known values of $a=-2.5m/s^2$, $v_f=0.0m/s$, and $v_o=13m/s$ yields a minimum distance of $$\Delta x=\frac{-(13m/s)^2}{2(-2.5m/s^2)}$$ $$\Delta x=34m$$
