Respuesta :
Answer:
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
Explanation:
We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+W_{disp}[/tex] (1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{disp}[/tex] - Work dissipated by friction, measured in joules.
By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2}) =\mu_{k}\cdot N\cdot \Delta s[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the skier, measured in meters.
[tex]N[/tex] - Normal force from the incline on the skier, measured in newtons.
[tex]\Delta s[/tex] - Distance covered by the skier, measured in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
The normal force exerted on the skier and the covered distance are, respectively:
[tex]N = m\cdot g\cdot \cos \theta[/tex] (3)
[tex]\Delta s = \frac{z_{1}-z_{2}}{\sin \theta}[/tex] (4)
Where [tex]\theta[/tex] is the angle of the incline above the horizontal, measured in sexagesimal degrees.
By applying (3) and (4) in (2), we get that:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot m\cdot g \cdot \cos \theta \cdot \left(\frac{z_{1}-z_{2}}{\sin \theta} \right)[/tex]
[tex]g\cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})= \mu_{k}\cdot g \cdot \left(\frac{z_{1}-z_{2}}{\tan \theta} \right)[/tex] (5)
Then, we clear the velocity of the skier at the bottom of the ramp is: ([tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.1[/tex], [tex]\theta = 40^{\circ}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}-z_{2} = 40\,m[/tex])
[tex]\left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]
[tex]2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = v_{1}^{2}-v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}-2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2})}[/tex] (6)
[tex]v_{2} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(\frac{0.1}{\tan 40^{\circ}} -1\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m)}[/tex]
[tex]v_{2} \approx 26.288\,\frac{m}{s}[/tex]
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
