Question

Runaway ski slides down a 260-m-long slope inclined at 36º with the horizontal. If the initial speed is 12.0 m/s, how long does it take the ski to reach the bottom of the incline if the coefficient of kinetic friction between the ski and snow is 0.100?

Answer 1

Given:
\(d=260 m\)
\(\theta=36^\circ\)
\(v_0=12.0 m/s\)
\(\mu_k=0.100\)

Looked for: t=? s (The time to reach the bottom of this 260-m slope )

Answer 2

The acceleration in the direction of the relative motion is given by:
\(\color{#000000}{ \displaystyle a=g(\sin \theta-\mu_k\cos \theta) }\)

Answer 3

So, therefore, velocity at time t, in the direction of the relative motion is:
(\(u\) is the initial speed, or the velocity in the direction of relative motion,
and \(a\) is the acceleration in the direction of relative motion)

\(\color{#000000}{ \displaystyle v(t)=u+at }\)
\(\color{#000000}{ \displaystyle v(t)=u+gt(\sin \theta-\mu_k\cos \theta) }\)

Answer 4

That means that the displacement in the direction of the relative motion would be the integral of this with respect to time. (\(\color{#000000}{ \displaystyle \Delta s }\) - distance, or, displacement in the direction of rel. mot. and \(\color{#000000}{ \displaystyle s_0 }\) is the initial displacement in the direction of the rel. mot.)

\(\color{#000000}{ \displaystyle \Delta s=ut+\frac{1}{2}gt^2(\sin \theta-\mu_k\cos \theta)+s_0 }\)

Answer 5

In this case:
\(\color{#000000}{ \displaystyle 260=12t+\frac{1}{2}(9.80)t^2(\sin 36-0.1\cos 36) }\)

(Assuming, that \(\color{#000000}{ \displaystyle s_0=0 }\) since no initial displacement is mentioned.)

Answer 6

So to find the time to reach the bottom, all I have to do is to solve this quadratic for t.

Answer 7

For that: \(\color{#000000}{ \displaystyle t\approx 8.10 }\)

So, if the coefficient of the Kinetic Friction is 0.150, then:
\(\color{#000000}{ \displaystyle 260=12t+\frac{1}{2}(9.80)t^2(\sin 36-0.15\cos 36) }\)

\(\color{#000000}{ \displaystyle t\approx 8.36 }\)