An extreme skier, starting from rest, coasts down a mountainthat makes an angle&nbsp;25.0∘&nbsp;with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 11.9 m before coming to the edge of a cliff. Without slowing down, she skis offthe cliff and lands down hill at a point whose vertical distance is 4.20 m below the edge. How fast is she going just before she lands?

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An extreme skier, starting from rest, coasts down a mountainthat makes an angle&amp;nbsp;25.0∘&amp;nbsp;with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 11.9 m before coming to the edge of a cliff. Without slowing down, she skis offthe cliff and lands down hill at a point whose vertical distance is 4.20 m below the edge. How fast is she going just before she lands?

falhiblesw · Accepted Answer

First find the skier&#039;s final speed when she clears the cliff: Before we can use the equation vf2=vo2+2ad we need her acceleration. If you drew a free body diagram, you should come up with thefollowing equation: mgsin⁡θ−Fiction=ma Fiction=mgcos⁡θ mgsin⁡θ−mgcos⁡θ=ma the masses cancel and you have: a=g(sin⁡θ−cos⁡θ)=9.81(sin⁡(25)−cos⁡(25)⋅0.2)=2.37ms2 Now use acceleration to find her final velocity before leavingthe cliff: vo=0 a=2.37 ms2 d=11.9m vf=(2)(9.81)(11.9)=7.51ms She leaves the cliff at this speed at 25 deg to the horizontal, so V⊗=7.51⋅cos⁡(25)=6.80ms Voy=7.51⋅sin⁡(25)=3.17 ms Now we just need to find Vfx and Vfv , and then take their magnitude to get the final answer. Vfx=6.8 ms Vfy2=Voy2+2ad=3.172+2(9.81)(4.2)=92.45 ms vf=vfx2+vfy2=6.82+92.45≈11.8 ms

Vasquez · Accepted Answer

Answer:10.62m/sExplanation:The initial mechanical energy is given by the sum of potential energy and kinetic energy:Einitial=mghi+12mvi2where m is the mass of the skier and g is the acceleration due to gravity.The final mechanical energy is given by the sum of potential energy and kinetic energy at the landing point:Efinal=mghf+12mvf2where vf is the final speed of the skier just before she lands.Since there is no change in mechanical energy between the initial and final states, we can equate the two expressions:mghi+12mvi2=mghf+12mvf2Canceling out the mass m from both sides and rearranging the equation, we get:ghi+12vi2=ghf+12vf2We can solve for the final speed vf:vf2=vi2+2g(hi−hf)To calculate the initial height hi and the height difference (hi−hf), we use trigonometry.hi=distance·sin(angle)hi=11.9m·sin(25.0∘)Now, the height difference (hi−hf) is equal to the vertical distance of the cliff:(hi−hf)=4.20mSubstituting these values into the equation for final speed vf, we get:vf=vi2+2g(hi−hf)vf=vi2+2·9.8m/s2·(11.9m·sin(25.0∘)−4.20m)Now we can calculate the final speed vf by plugging in the given values and solving the equation.(11.9m·sin(25.0∘)−4.20m)≈5.77mPlugging this value back into the equation for vf:vf=02+2·9.8m/s2·5.77mCalculating the term inside the square root:2·9.8m/s2·5.77m≈112.61m2/s2Finally, taking the square root to find vf:vf≈112.61m2/s2≈10.62m/sTherefore, the skier&#039;s final speed just before she lands is approximately 10.62m/s.

Don Sumner · Accepted Answer

Step 1:To solve this problem, we can use the principles of conservation of energy. The initial mechanical energy of the skier at the top of the hill is converted into kinetic energy as she coasts down the mountain and potential energy as she falls off the cliff.Let&#039;s denote the initial height of the skier at the top of the hill as hi, the final height at the point where she lands as hf, and the distance she coasts down the mountain as d. The skier starts from rest, so her initial kinetic energy is zero.The initial potential energy at the top of the hill is given by:PEi=mghiwhere m is the mass of the skier, g is the acceleration due to gravity, and hi is the initial height. The initial potential energy is converted into kinetic energy as she coasts down the mountain:KEi=12mvi2where vi is the initial velocity of the skier.The work done against friction as she coasts down the mountain is given by:Wfriction=μk·m·g·dwhere μk is the coefficient of kinetic friction, m is the mass of the skier, g is the acceleration due to gravity, and d is the distance she coasts.The final potential energy at the point where she lands is given by:PEf=mghfwhere hf is the final height.The final kinetic energy just before she lands is given by:KEf=12mvf2where vf is the final velocity of the skier.According to the conservation of energy principle, the initial mechanical energy is equal to the final mechanical energy:PEi+KEi−Wfriction=PEf+KEfPlugging in the respective values, we get:mghi+12mvi2−μk·m·g·d=mghf+12mvf2Step 2:Now we can solve for the final velocity vf just before the skier lands. Rearranging the equation, we have:12mvf2=mghi+12mvi2−μk·m·g·d−mghfDividing both sides by m and rearranging again, we obtain:vf2=2ghi+vi2−2μk·g·d−2ghfFinally, taking the square root of both sides, we find the expression for the final velocity vf:vf=2ghi+vi2−2μk·g·d−2ghfNow we can substitute the given values into the equation to calculate the final velocity vf.

nick1337 · Accepted Answer

Let&#039;s denote the speed of the skier just before she lands as v. The initial potential energy of the skier is equal to her mass (m) multiplied by the acceleration due to gravity (g) and the initial vertical displacement (h1):Initial&amp;nbsp;potential&amp;nbsp;energy=mgh1Here, h1 is the vertical distance from the starting point to the edge of the cliff.The final potential energy of the skier just before she lands is given by:Final&amp;nbsp;potential&amp;nbsp;energy=mgh2 where h2 is the vertical distance from the landing point to the edge of the cliff. Since the skier lands downhill, h2 is negative.The loss in potential energy is equal to the gain in kinetic energy. The loss in potential energy can be calculated by subtracting the final potential energy from the initial potential energy:Loss&amp;nbsp;in&amp;nbsp;potential&amp;nbsp;energy=Initial&amp;nbsp;potential&amp;nbsp;energy−Final&amp;nbsp;potential&amp;nbsp;energyThis loss in potential energy is equal to the gain in kinetic energy:Loss&amp;nbsp;in&amp;nbsp;potential&amp;nbsp;energy=12mv2Now we can equate the two expressions for the loss in potential energy:mgh1−mgh2=12mv2We can cancel out the mass (m) from both sides of the equation:gh1−gh2=12v2We know the acceleration due to gravity (g) and the vertical distances h1 and h2, so we can substitute their values into the equation and solve for v:9.8m/s2×h1−9.8m/s2×(−4.2m)=12v29.8m/s2×(h1+4.2m)=12v2v2=19.6m/s2×(h1+4.2m)Finally, we can take the square root of both sides to find the speed of the skier just before she lands:v=19.6m/s2×(h1+4.2m)Substituting the values:v=19.6m/s2×(4.98m+4.2m)Calculating v:v≈19.6m/s2×9.18mv≈178.888m2/s2v≈13.37m/sTherefore, the skier is going approximately 13.37m/s just before she lands.

An extreme skier, starting from rest, coasts down a mountainthat makes an angle 25.0^\circ with the horizontal. The coefficient of kinetic friction be

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