Question

a skier starts from rest at the top of a 45.0 m hill, coasts down a 30° slope into a valley and continues up to the top of a 40.0 m hill. both hill heights are measured from the valley floor. assume the skier puts no effort into the motion (they simply coast) and there is no friction. how fast will the skier be moving on the top of the 40.0 m hill? a 3.6 m/s b 4.6 m/s c 9.9 m/s d 2.5 m/s

Explanation:

Step1: Apply conservation of mechanical energy

The initial mechanical energy \(E_1\) is gravitational potential energy \(E_{p1}=mgh_1\) (since the skier starts from rest, \(E_{k1} = 0\)). The final mechanical energy \(E_2\) is the sum of gravitational potential energy \(E_{p2}=mgh_2\) and kinetic energy \(E_{k2}=\frac{1}{2}mv^{2}\). By the conservation of mechanical energy \(E_1 = E_2\), so \(mgh_1=mgh_2+\frac{1}{2}mv^{2}\).

Step2: Simplify the equation

Divide the entire equation \(mgh_1=mgh_2+\frac{1}{2}mv^{2}\) by \(m\) (mass \(m\) cancels out). We get \(gh_1=gh_2+\frac{1}{2}v^{2}\). Rearrange for \(v\): \(\frac{1}{2}v^{2}=g(h_1 - h_2)\). Given \(g = 9.8\space m/s^{2}\), \(h_1=45.0\space m\), \(h_2 = 40.0\space m\). Substitute the values: \(\frac{1}{2}v^{2}=9.8\times(45 - 40)\).

Step3: Solve for \(v\)

First, calculate \(9.8\times(45 - 40)=9.8\times5 = 49\). Then from \(\frac{1}{2}v^{2}=49\), multiply both sides by 2 to get \(v^{2}=98\). Take the square - root: \(v=\sqrt{98}\approx9.9\space m/s\).

Answer:

C. \(9.9\space m/s\)