question about kinetic friction?

Question

A 63 kg skier coasts up a snow covered hill that makes an angle of 25 degrees with the horizontal. The intial speed is 6.6 m/s, she coasts a distance of 1.9 m, the final speed is 4.4 m/s.
A) find the work done by the kinetic frictional force that acts on the skis.

B) what is the magnitude of the kinetic frictional force?

Answer 1

well...
A) I believe you do:
V^2=Vo^2+2a(x-x0)
where V=6.6m/s
Vo=4.4 m/s
x-xo= 1.9
then solve for a

Free body diagram... x-axis on the slope... gravity and friction
Sum the forces in the x and you get mgsin(25)+f=ma then plug in values to get f... work done by frictional force (thermal energy) is equal to f*d

Magnitude of Kinetic frictional force is i believe is the value of f in the equation

Answer 2

Part B is easier, so let's answer it first.

First, determine the total acceleration on the skier:
Vf² - Vi² = 2ad
a = (Vf² - Vi²)/2d = (4.4² - 6.6²)/(2 x 1.9)
a = -6.37 m/sec²

This acceleration is a combination of gravity and friction. How much acceleration is due to gravity? On a 25° slope, we are interested in the component of gravity parallel to the surface, so the acceleration due to gravity is:
g = (-9.8 m/sec²)sin(25°) = -4.14 m/sec²

Therefore the acceleration due to friction is
(-6.37) - (-4.14) = -2.23 m/sec²

We know that F = ma, m = 63 kg and a = -2.23 m/sec², so
F = (63)(-2.23)kg-m/sec² = -140 N (the minus sign is irrelevant)

Now to the answer to part A.

Work is force x distance
therefore the work of the friction is
140N x 1.9m = 266 Nm = 266J

Answer 3

i dont know have u tried searching google