Physics Question involving a skier and a snowdrift....

[Serella_Madole]

Homework Statement

[/B]

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.1 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Homework Equations

TE(initial) = TE(final)

The Attempt at a Solution

TEi = TEf
Kei + Pei = Elastic Energy + Kef (Pei = zero)
1/2mv^2 =1/2kx^2
k = mv^2/x^2
k = 14583.3

F=-kx^2
F=64312.353N

This is not correct; the correct answer is 4210N. Apparently you can use the Work Energy Theorem in order to determine this value. I am not sure how though. Please assist in helping me understand. [/B]

 

[SteamKing]

Homework Statement

[/B]

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.1 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Homework Equations

TE(initial) = TE(final)

The Attempt at a Solution

TEi = TEf
Kei + Pei = Elastic Energy + Kef (Pei = zero)
1/2mv^2 =1/2kx^2
k = mv^2/x^2
k = 14583.3

F=-kx^2
F=64312.353N

This is not correct; the correct answer is 4210N. Apparently you can use the Work Energy Theorem in order to determine this value. I am not sure how though. Please assist in helping me understand. [/B]

Is there more information given with this problem? Like the velocity of the skier before she hits the snowdrift?

 

[Serella_Madole]

Is there more information given with this problem? Like the velocity of the skier before she hits the snowdrift?

I had to determine the final velocity on my own which is 17.8 m/s. The initial velocity is zero based on the first part of the question. "A 61.0-kg skier starts from rest at the top of a ski slope of height 70.0 m."

 

[SteamKing]

We can't help you if you don't give us all the information at your disposal.

 

[Serella_Madole]

We can't help you if you don't give us all the information at your disposal.

Sorry for the late response!

Here is all the information accompanying the question!

Part A
If frictional forces do −1.05×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be g = 9.80 m/s2.

v = 32.1 m/s

Part B
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.25. If the patch is of width 70.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?

v = 17.8 m/s

I hope this helps!

 

[haruspex]

1/2mv^2 =1/2kx^2

A snowdrift does not behave like a spring.

F=-kx^2

Even if it did behave like a spring, that is not the right equation for the force exerted by a spring.

Unfortunately, the question itself is flawed. "Average force" means the average over time. Since you do not know how long the skier took to stop, there is no way to work out the average force. In order to get the book answer, you will have treat the force as constant.
See "3. Average force" at https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/