Question

A skier starts at the top of a frictionless slope and pushes off with a speed of 3.0 m/s. The elevation of the slope is 40 m. She skis down the slope to a valley with elevation 0.0 m and then glides to the peak of an adjacent slope that is at an elevation of 25 m. Calculate her speed at the second peak.

Answer 1

To calculate the skier's speed at the second peak, we need to apply the principle of conservation of mechanical energy. The skier's speed at the second peak is approximately 5.1 m/s. Conservation of mechanical energy equation: ΔPE + ΔKE = 0.

Step-by-step explanation:

To calculate the skier's speed at the second peak, we need to apply the principle of conservation of mechanical energy. Since the slope is frictionless, the only forces acting on the skier are the gravitational force and the normal force. Initially, the skier has gravitational potential energy due to the elevation of the slope, and this energy gets converted into kinetic energy as the skier moves down the slope.

We can use the conservation of mechanical energy equation: ΔPE + ΔKE = 0, where ΔPE is the change in gravitational potential energy and ΔKE is the change in kinetic energy.

At the first peak (elevation = 40 m), the skier has potential energy and no kinetic energy.

At the second peak (elevation = 25 m), the skier has no potential energy and some kinetic energy. The change in potential energy is ΔPE = 40 m - 25 m = 15 m.

Since there is no friction, the total mechanical energy is conserved, so we can write the equation as:

mgh1 + 1/2 mV1^2 = mgh2 + 1/2 mV2^2, where h1 = 40 m, h2 = 25 m, V1 = 3.0 m/s (initial speed), and V2 is the speed at the second peak.

Simplifying the equation, we get: 40 * 9.8 + 1/2 * 3.0^2 = 25 * 9.8 + 1/2 * V2^2. Solving for V2, we find that the skier's speed at the second peak is approximately 5.1 m/s.

Answer 2

Given:

The speed of the skier at the top of the first peak is,

`v_1=3.0\text{ m/s}`

The elevation of the slope is,

`h_1=40\text{ m}`

She skis down the slope to a valley with an elevation of 0.0 m and then glides to the peak of an adjacent slope that is at an elevation of

`h_2=25\text{ m}`

To find:

The speed at the second peak

Explanation:

The diagram can be depicted below as:

Using the conservation of the mechanical energy, we can write,

The total energy at peak 1 = The total energy at peak 2

So,

`\begin{gathered} mgh_1+(1)/(2)mv_1^2=mgh_2+(1)/(2)mv_2^2 \\ v_2^2=2gh_1+v_1^2-2gh_2 \\ v_2=√(2gh_1+v_1^2-2gh_2) \end{gathered}`

Substituting the values we get,

`\begin{gathered} v_2=√(2*9.8*40+(3.0)^2-2*9.8*25) \\ =√(303) \\ =17.4\text{ m/s} \end{gathered}`

Hence, the speed at the second peak is 17.4 m/s.