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## Homework Statement

A 15.0 kg stone slides down a snow-covered hill (the figure ), leaving point A with a speed of 10.0 m/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 100m and then runs into a very long, light spring with force constant 2.50 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

Here is the picture: http://session.masteringphysics.com/problemAsset/1260235/2/YF-07-34.jpg

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

B) How far will the stone compress the spring?

## Homework Equations

PE = mgh

KE = .5mv^2

spring = .5kx^2

## The Attempt at a Solution

A) I did this:

PE = (15.0)(9.80)(20.0) = 2943 J

KE= .5(15.0)(10.0)^2 = 750 J

2943+750 = 3693 J

3693 = .5(15.0)(v)^2-.5(15.0)(10.0)^2

v=19.80 m/s

B) work done by spring = .5(2.50)(x)^2 = 1.25x^2

work done by friction = (0.2)(15.0)(9.8)x = 29.4x

Using quadratic:

1.25x^2 + 29.43x - 3693 = 0

x= 43.84 or -67.38

x= only 43.84 meters.

Do these answers seem right? thanks!

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