Skier skies down slope, including friction, calculate final speed

[wiggle]

Homework Statement

A 50Kg skier skies down a 25 degree slope. at the top of the slope her speed is 4m/s and accelerates down the hill. the coefficient of friction is 0.12 between skies and snow. ignoring air resistance calculate her speed at point that is displaced 60m downhill.

Homework Equations

E=KE+PE
E=.5*m*V^2+m*g*h

The Attempt at a Solution

I believe I solved it correctly using E0=.5(50)4^2+50(9.8)(60sin(25)) and getting 12650

then setting 12650 = .5*m*Vf^2 and I'm getting Vf=22.5 m/s

However I'm unsure how to go about solving while taking into consideration the friction, since there is a non conservative force at work using KE0+PE0=KEf+PEf would get me close but not all the way there.

 

[Doc Al]

The work done by friction will reduce the total mechanical energy.

 

[wiggle]

Right.

So I've just broken it up into x and y components

Fx=Wsin25-.12(Wcos(25))=m*a

I calculated for acceleration in the x plane, and found a=3.08 m/s^2

Then I just did

Vf=sqr(2ax+V0^2)
sqr(2(3.08)(60)+4^2)= Vf= 19.6 m/s

Does this seem correct?

 

[Doc Al]

Looks good to me.

 

[Nickie]

To solve for the final speed of the skier taking into account friction, we need to use the work-energy theorem. This theorem states that the net work done on an object is equal to the change in kinetic energy of the object. In this case, the net work done on the skier is equal to the work done by the gravitational force minus the work done by the frictional force. This can be expressed as:

Wnet = Wg - Wf

where Wg is the work done by the gravitational force and Wf is the work done by the frictional force. The work done by the gravitational force can be calculated using the equation Wg = mgh, where m is the mass of the skier, g is the acceleration due to gravity, and h is the height difference between the top and bottom of the slope.

The work done by the frictional force can be calculated using the equation Wf = μmgd, where μ is the coefficient of friction, m is the mass of the skier, g is the acceleration due to gravity, and d is the distance traveled downhill.

Therefore, the net work done on the skier can be expressed as:

Wnet = mgh - μmgd

Using the work-energy theorem, we can equate this to the change in kinetic energy of the skier:

Wnet = ΔKE = KEf - KE0

Substituting the values for Wnet, mgh, and μmgd, we get:

mgh - μmgd = 1/2mvf^2 - 1/2mv0^2

where vf is the final speed and v0 is the initial speed.

Solving for vf, we get:

vf = sqrt(2gh - 2μgd + v0^2)

Substituting the given values, we get:

vf = sqrt(2(9.8)(60sin25) - 2(0.12)(50)(9.8)(60) + 4^2) = 20.6 m/s

Therefore, the final speed of the skier taking into account friction is 20.6 m/s. This is slightly lower than the calculated speed of 22.5 m/s when ignoring friction. This shows that friction does have a significant impact on the speed of the skier and should be taken into consideration in calculations.