Question

1. Once again we have a skier on an inclined plane. The skier has mass M and starts from rest. Her speed at the bottom of the slope is 12.0 meters/second. The hill is inclined 30 degrees from the horizontal and the length of the slope is 100.0 meters. Use the law of conservation of energy to determine her final speed and write that number down here: What could account for the difference in your calculated value using conservation of energy, and the given value? (hint: is the hill completely frictionless)? What is the coefficient of kinetic friction of the slope? Some ideas to think about: 1. We can no longer calculate the total energy at two different points and then set those values equal to each other. This is because the total energy is not transferred between the kinetic and potential energy; some of the energy is "lost" or transferred as heat. Friction is a dissipative force and is not conservative. 2. Clearly the skier's kinetic energy is changing; she starts from rest and ends with a speed of 12 meter/second. 3. Use the work energy theorem find the coefficient of kinetic friction. (Refer back to the problem we did in class last meeting. How did we handle the rough patch that the skier encountered at the bottom of the hill).

please explain and write neatly

Answer 1

- Sno- h halsno=100&n 30-50 1 BX n e =100 7 Using conservation of Energy sno=4 halsino = 1006m 30-50 Potential Enerzy = kenetic energy. migh= 1 mv2 v ² - gh v=J2gh V- 1289.8*50 V> 31. 30 ms. 2) Some of energy is lost as work done by fraction which is negative. 3) AKE + APE = Wenchon N txm xv²-0 + (0-mgh) = 4 mg coso x la cosisoo - friction (12) - 9.8*50 = 4x9.8 x cosso x 100 x-1 to my sno mgaso — 418 = -848.704 u = 418 0.4925 Namycoso frichun. HN 848.700-4928 Scanned with CamScanner