Homework Help: Physics

Posted by Alex on Saturday, December 15, 2012 at 4:59pm.

A 59-kg person on skis starts from rest down a hill sloped at 36° from the horizontal. The coefficient of friction between the skis and the snow is 0.16. After the skier had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero.

What is the new coefficient of friction?

How fast would the skier now be going after skiing for another 5.0 s?

Physics - Elena, Saturday, December 15, 2012 at 5:33pm
ma=mgsinα –F(fr) = mgsinα – μmgcosα
a= g(sinα – μcosα ) = 9.8• (0.59-0.16•0.81)=4.5 m/s²
The speed after the first 5 sec is
v=at= 4.5•5=22.5 m/s
If the net force=0, a=0 =>
0 = mgsinα – μ₁•mgcosα ,
μ₁ =sinα/cosα=tanα=0.73.
If a=0 => v=const=22.5 m/s

Answer this Question

Related Questions

physics - A 69 kg person on skis is going down a hill sloped at 36° below ...
science - A 75-kg person on skis is going down a hill sloped at 30.0°. The ...
physics - A 75g person on skis is going down a hill sloped at an angle 30.the ...
Physics - A 61-kg person on skis is going down a hill sloped at 30¡Æ...
Pls help me now - A 75g person on skis is going down a hill sloped at an angle ...
Physics - A skier starts from rest at the top of a hill that is inclined at 9.8&...
physics - A skier starts from rest at the top of a hill that is inclined at 9.5&...
physics - A skier starts from rest at the top of a hill that is inclined at 9.5&...
physics - A skier starts from rest at the top of a hill that is inclined at 10.3...
physics...again :/ - A skier starts from rest at the top of a hill that is ...

For Further Reading

First Name:
School Subject:
Answer: