A skier whose mass is 70 kg, stands at the topof a 10 degree slope on his new fr

Question

A skier whose mass is 70 kg, stands at the topof a 10 degree slope on his new frictionless skies. A stronghorizontal wind blows against him with a force of 50N. Explain howyou find the skiers speed after traveling some distance down theslope, without using Newton's second law. What is the skiers speedafter traveling100 meters down the slope? F=mgSin 10 degrees _ (50/Sin 10degrees) F=70*9.81*Sin(10) - (50/Sin 10 degrees) F=119.24 - 50.78 F=68.46N W=(F Cos )* S Note:Cos = 0 W=68.46 * 100 W=6834J W=1/2 mvf2 Is this the right approach? A skier whose mass is 70 kg, stands at the topof a 10 degree slope on his new frictionless skies. A stronghorizontal wind blows against him with a force of 50N. Explain howyou find the skiers speed after traveling some distance down theslope, without using Newton's second law. What is the skiers speedafter traveling100 meters down the slope? F=mgSin 10 degrees _ (50/Sin 10degrees) F=70*9.81*Sin(10) - (50/Sin 10 degrees) F=119.24 - 50.78 F=68.46N W=(F Cos )* S Note:Cos = 0 W=68.46 * 100 W=6834J W=1/2 mvf2 Is this the right approach? Is this the right approach?

Explanation / Answer

Given : F=mgSin 10degrees _ (50/Sin 10 degrees) F=70*9.81*Sin(10) - (50/Sin 10degrees) F=119.24 - 50.78 F=68.46N W=(F Cos )* S Note:Cos = 0 W=68.46 * 100 W=6834J W=1/2mvf2 Yes , your appoch is correct I hope it helps you F=mgSin 10degrees _ (50/Sin 10 degrees) F=70*9.81*Sin(10) - (50/Sin 10degrees) F=119.24 - 50.78 F=68.46N W=(F Cos )* S Note:Cos = 0 W=68.46 * 100 W=6834J W=1/2mvf2 Yes , your appoch is correct I hope it helps you

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