The coefficient of kinetic friction for a 22-kg bobsled on a track is 0.10. What force is required to push it down along a 6.0° incline and achieve a speed of 60 km/h at the end of 75 m?

Given data:

The mass of the bobsled is $m=22\mathrm{kg}$.

The angle of incline is $\theta =6.0°$.

The coefficient of kinetic friction is ${\mu }_{\mathrm{k}}=0.10$.

Initial speed of the bobsled is $v_{\mathrm{i}}=0$.

The final speed of the bobsled is $v_{\mathrm{f}}=60\mathrm{km}/\mathrm{h}=16.67\mathrm{m}/\mathrm{s}$.

The distance moved on the plane is $\Delta x=75\mathrm{m}$.

Let the force required to push the bobsled be F.The bobsled moves down, and the frictional force on it acts upward.

The normal force acting on the bobsled is $N=mg\cos \theta$(if you balance the force perpendicular to the plane).

The frictional force on the bobsled is:

$\begin{array}{c}{f}_{\mathrm{k}}={\mu }_{\mathrm{k}}N\\ ={\mu }_{\mathrm{k}}mg\cos \theta \end{array}$

The net force on the bobsled is:

$\begin{array}{c}{F}_{\mathrm{net}}=F+mg\sin \theta -f_{\mathrm{k}}\\ =F+mg\sin \theta -{\mu }_{\mathrm{k}}mg\cos \theta \\ =F+mg\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)\end{array}$

Therefore, the acceleration of the bobsled is:

$\begin{array}{c}a=\frac{F_{\mathrm{net}}}{m}\\ =\frac{F+mg\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)}{m}\\ =\frac{F}{m}+g\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)\end{array}$

Now, from the equation of motion, you can write:

$\begin{array}{c}{v}_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2a\Delta x\\ v_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2\left(\frac{F}{m}+g\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)\right)\Delta x\\ {\left(16.67\mathrm{m}/\mathrm{s}\right)}^2=0^2+\left[2\times \left(\frac{F}{22\mathrm{kg}}+\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(\sin 6.0°-0.10\cos 6.0°\right)\right)\times 75\mathrm{m}\right]\\ F=39.66\mathrm{N}\\ F\approx 4.0\times {10}^1\mathrm{N}\end{array}$

Hence, the value of the required force is $4.0\times {10}^1\mathrm{N}$.