What is the final position of the stone after being stopped by the spring?

[lgn_barnard]

Homework Statement

A 15.0kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point A with a speed of 11.0m/s . There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.10N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

What is the speed of the stone when it reaches point B?

22.65 m/s (this is correct)

How far will the stone compress the spring?

Will the stone move again after it has been stopped by the spring?

picture: Mastering Physics

Homework Equations

PE = mgh
KE = .5mv^2
spring energy = .5kx^2

The Attempt at a Solution

I set up the equation

.5mv^2=.5kx^2+(kinetic friction)mgL

0.5(15)(22.65^2)=0.5(2.1)x^2 + 0.2(15)(9.8)(100)

I solved for x to equal 29.4, but this is wrong

 

[Simon Bridge]

Please show your reasoning.
Please attach any diagrams referred to.

What kinds of energy does the stone start out with?
It looks like you have included only the initial kinetic energy - but with the wrong speed.

 

[Legaldose]

First find the final velocity of the stone when it reaches the spring. From this you can find the kinetic energy, this will be the energy that it imparts on the spring. Now once it's stopped find the static frictional force on the stone, and compare it with the force that the spring is exerting. This is how I would solve it, break it into bite sized chunks.

 

[lep11]

Your equations look fine. However, you'll have to take into account that the friction force acts also during the compression of the spring. It is only said that the stone travels 100m before it runs into the spring.

 

[lgn_barnard]

Your equations look fine. However, you'll have to take into account that the friction force acts also during the compression of the spring. It is only said that the stone travels 100m before it runs into the spring.

So if I add (.2)(15)(9.8)(x) to the right side of the equation if should come out correct?

 

[Simon Bridge]

Why don't you try it and see?
It's faster that way, and you learn more.

is 22.65m/s the speed when the stone hits the friction section?
[edit - oh I see it is]

we don't have the diagram remember.

It is best practice to do all the algebra before plugging numbers in.