Calculating Work Done by Kinetic Frictional Force on a Coasting Skier

[workhard]

A 67.9kg skier coasts up a snow-covered hill that makes an angle of 27.7o with the horizontal. The initial speed of the skier is 8.54m/s. After coasting a distance of 1.94m up the slope, the speed of the skier is 3.15m/s. Calculate the work done by the kinetic frictional force that acts on the skis.

This is a problem that I seem to keep getting wrong. I have spent over an hour on it now...

so far I have come up with that i will find my initial and final KE by .5mv^2

How do I incorporate the conservational equation into this?

My mind is jumbled

Thanks.

 

[Chi Meson]

change in KE = (-) change in PE + Q. (Q is heat generated through work done by friction)

 

[kyle8921]

I can offer some guidance on how to approach this problem. First, let's define the variables we will be using:

m = mass of the skier (67.9kg)
θ = angle of the slope (27.7°)
vi = initial velocity (8.54m/s)
vf = final velocity (3.15m/s)
d = distance traveled up the slope (1.94m)
μk = coefficient of kinetic friction

To calculate the work done by the kinetic frictional force, we need to use the formula W = Fd, where W is the work done, F is the force, and d is the distance over which the force acts. In this case, the force we are interested in is the kinetic frictional force, which can be calculated using the formula Fk = μkmgcosθ, where μk is the coefficient of kinetic friction, m is the mass of the skier, g is the acceleration due to gravity, and cosθ accounts for the angle of the slope.

Now, to incorporate the conservation of energy equation, we can use the fact that the total work done on the skier is equal to the change in kinetic energy. In other words, Wtotal = ΔKE. We can also break this down into two parts: the work done by the kinetic frictional force (Wk) and the work done by the gravitational force (Wg). Therefore, we can write the equation as Wk + Wg = ΔKE.

Since the skier is coasting up a slope, the work done by the gravitational force is negative, as it is acting in the opposite direction of motion. This means that Wg = -mgh, where h is the vertical distance traveled, which in this case is equal to dsinθ. Therefore, Wg = -mgsinθd.

Now, we can substitute these values into our equation and solve for Wk:

Wk + (-mgsinθd) = ½ mvf^2 – ½ mvi^2

Wk = ½ mvf^2 – ½ mvi^2 + mgsinθd

Wk = ½ (67.9kg)(3.15m/s)^2 – ½ (67.9kg)(8.54m/s)^2 + (67.9kg)(9.8m/s^2)sin(27.7