Question

A 60 kg skier leaves the end of a ski-jump ramp with a velocity of 24 m/s directed $\mathbf{25}\mathbf{°}$above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 22 m/s, landing 14 m vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

Short answer

The reduction in the mechanical energy of the skier-earth system is $1.1\times {10}^4\mathrm{J}$.

Step-by-step solution

Step 1: Given Data

The mass of the skier is 60 kg.

The speed of the skier when it leaves the ramp is 24 m/s.

The angle made by the ramp is$25°$above the ground.

The vertical distance travelled by skier below the ramp is 14 m.

The speed of the skier as it reaches the ground is 22 m/s.

Determining the concept

The total energy of the skier – earth system is conserved. The mechanical energy of the skier at the starting point is reduced as he reaches the ground. The reduction in the energy is the amount of energy lost during the motion due to air drag.

Formulae are as follow:

Potential energy,$PE=mgh$

$KE=\frac{1}{2}m{v}^2$

Energy loss = change in PE + change in KE

Where, KE is kinetic energy, PEis potential energy, m is mass, v is velocity, g is an acceleration due to gravity and h is height.

Determining the reduction in the mechanical energy of the skier-earth system

Consider, the level of the ramp from which the skier jumps as level zero. Hence, the ground level at which the skier lands will be taken as negative. The initial mechanical energy of the skier –earth system is lost due to air drag as he reaches the ground. This reduction in energy is calculated as,

$$\begin{gathered} ∆E=∆PE+∆KE \\ =\left(mg{h}_i-mg{h}_f\right)+\left(\frac{1}{2}m{v}_i^2-\frac{1}{2}m{v}_f^2\right) \\ ∆E=\left(60\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\left(14\mathrm{m}-0\right)\right)+\left(\frac{1}{2}\times 60\mathrm{kg}\times \left({24}^{2.}{\left(\mathrm{m}/\mathrm{s}\right)}^2-{22}^2{\left(\mathrm{m}/\mathrm{s}\right)}^2\right)\right) \\ ∆\mathrm{E}=8232+2760 \\ =1.1\times {10}^4\mathrm{J} \end{gathered}$$

The negative sign indicates the change in energy is the loss of the energy.

Hence,the reduction in the mechanical energy of the skier-earth system is$1.1\times {10}^4\mathrm{J}$.

Therefore, the law of conservation of energy holds true for the skier-earth system. The mechanical energy at the start of the ramp is the sum of the mechanical energy at the end of the travel and the energy lost due to air drag.