91Ó°ÊÓ

Given data:

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

The angle of incline is $\mathrm{θ}=6.0°$.

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

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 $\mathrm{Δ}x=75\mathrm{m}$.

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 $N=mg\cos \mathrm{θ}$(if you balance the force perpendicular to the plane).

The frictional force on the bobsled is:

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

The net force on the bobsled is:

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

Therefore, the acceleration of the bobsled is:

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

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

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

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