## A community for students. Sign up today

Here's the question you clicked on:

## unimatix one year ago 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?

• This Question is Closed
1. anonymous
2. unimatix

Thank you!

3. anonymous

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

4. unimatix

Where did you get those numbers from?

5. IrishBoy123

|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 $$\mu$$

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy