Sam, whose mass is 71.0 kg , takes off down a 51.0 m high, 11.0 ∘ slope on his jet-powered skis. The skis have a thrust of 210 N . Sam's speed at the bottom is 41.0 m/s . What is the coefficient of kinetic friction of his skis on the snow?

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Sam, whose mass is 71.0 kg , takes off down a 51.0 m high, 11.0 ∘ slope on his jet-powered skis. The skis have a thrust of 210 N . Sam's speed at the bottom is 41.0 m/s . What is the coefficient of kinetic friction of his skis on the snow?

Mathematics
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http://www.jiskha.com/search/index.cgi?query=Sam%2C+whose+mass+is+71.0+kg+%2C+takes+off+down+a+51.0+m+high%2C+11.0+%26%238728%3B+slope+on+his+jet-powered+skis.+The+skis+have+a+thrust+of+210+N+.+Sam's+speed+at+the+bottom
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accelaration of sam=[(45^2).sin12]/(2*54)=A mass=65=M kinetic fric.=K gravity=g MAsin12-K(Mgcos12)=150 solve this then ull get K

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Where did you get those numbers from?
|dw:1443794291333:dw| resolving forces down the plane we have \(\large 210 + mg \sin 11 - \mu mg \cos 11 = ma\) where \(\mu \) is the coefficient of friction, m = 71kg, g is gravity, a = acceleration we know that acceleration a from \(\large v^2 = u^2 + 2ax\) ie \(\large 41^2 = 0^2 + 2 \;a (267.3)\). 267.3 is the length of the slope, you have all you need to get its dimensions from trig. put that together and solve for \(\mu\)

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