Question

A girl is sledding down a slope that is inclined at 30.0° with respect to the horizontal. The wind is aiding the motion by providing asteady force of 159. N that is parallel to the motion of the sled. The combined mass of the girl and the sled is 58.8 kg, and thecoefficient of kinetic friction between the snow and the runners of the sled is 0.113. How much time is required for the sled to traveldown a 223-m slope, starting from rest?

Answer 1

Let's draw the free body diagram:

Using newton's second law:

$\begin{gathered} \Sigma Fy=0 \\ so: \\ Wcos(30)-N=0 \\ so: \\ mgcos(30)=N \end{gathered}$$\begin{gathered} \Sigma Fx=ma \\ F+mgsin(30)-Ff=ma \\ a=\frac{F+mgsin(30)-Ff}{m} \end{gathered}$

We can solve for a using the values provided by the problem:

$\begin{gathered} a=\frac{159+58.8(9.8)sin(30)-\mu N}{m} \\ a=\frac{159+58.8(9.8)s\imaginaryI n(30)-(0.113)(58.8)(9.8)cos(30)}{58.8} \\ so: \\ a\approx6.645m/s^2 \end{gathered}$

Now, we can use the following kinematic equation:

$\begin{gathered} d=\frac{1}{2}at^2 \\ t^2=\frac{2d}{a} \\ t=\sqrt{\frac{2d}{a}} \\ t=\sqrt{\frac{2(223)}{6.645}} \\ t\approx8.19s \end{gathered}$

Answer:

8.19 seconds