A 68.0kg skier approaches the foot of a hill with a speed of 13.0m/s. The surfac

Question

A 68.0kg skier approaches the foot of a hill with a speed of 13.0m/s. The surface of this hill slopes up at 45.0degree above the horizontal and has coefficients of static and kinetic friction of 0.750 and 0.250, respectively, with the skis.

A small 0.140kg metal ball is tied to a very light (essentially massless) string 0.800m long to form a pendulum that is then set swinging by releasing the ball from rest when the string makes a 50.0degree angle with the vertical. Air drag and other forms of friction are negligible.

What is the tension in the string at that instant?

A 75.0 kg mountain climber is holding his 60.0 kg partner over a cliff when he suddenly steps on the ice with a coefficient of kinetic friction of 0.200 at the horizontal top of the cliff, as shown in the accompanying figure.(Figure 1) The rope has negligible mass and is held horizontally by the climber. There is no appreciable friction at the icy edge of the cliff. Use energy methods to calculate the speed of the climbers after the lower one has descended 1.80m starting from rest. What would be the speed of the climbers after they had moved 1.80m?

A 75.0 kgmountain climber is holding his 60.0kg partner over a cliff when he suddenly steps on frictionless ice at the horizontal top of the cliff, as shown in the accompanying figure.(Figure 1) The rope has negligible mass and is held horizontally by the climber. There is no appreciable friction at the icy edge of the cliff.

Use energy methods to calculate the speed of the climbers after the lower one has descended 1.60m starting from rest

Explanation / Answer

since there is friction present in the system I decided to use the work-energy theorem, thus making: Ef = Ei + W, or mgh = (1/2)mv^2 + W

Call the height that you are looking for y. The displacement along the hillside (for purposes of calculating work) is s = y / sin40o. So if you use the work energy theorem, you will have an equation involving only one unknown, y.

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