A 15.0 kg stone slides down a snow-covered hill leaving point A (at the top of the hill) with a speed of 10.0 m/s. There is no friction on he hill between point A and point B (at the foot of the hill). There is friction on the level ground at the bottom of the hill between point B and a wall some distance away. After entering the rough horizontal region, the stone travels 100 m and the runs into a very long light spring with force constant 2.00 N/m.
The coefficients of kinetic and static friction are 0.20 and 0.80, respectively.
a) What is the speed of the stone when it reaches point B ?
b) How far will the stone compress the spring ?
c) Will the stone move again after it has been stopped by the spring ?
E = K + U
K1 + Ugr = K2 + Ugr2
K1 + U1 + W (friction) = K2 + U2
K1 + Uel1 + W(friction) = K2 +Uel2
Where K1, K2 are initial and final Kinetic Energy,
U1, U2 are initial and final Potential Energy
Ugr1, Ugr2 initial and final Gravitational Potential Energy
Uel1, Uel2 initial and final Elastic potential Energy
W(friction) = work done by Friction (= (f) x (distance))
Uel = 1/2 k x(sqrd)
The Attempt at a Solution
Now I solved part a) with little difficulty noting that since this area has no friction then the stone's gravitational potential energy is entirely converted into kinetic energy by the time it reaches ground-level , so
mass x gravity x height = 1/2 m v(sqrd) (we also have a KE value for initial speed) ... the speed = 22.2 m/s .. which agrees with the book.
However the stumbling block is part b) .. no matter what I do I cannot get the correct answer for this part of the problem.
I attempted it as follows ..
I took it in 2 parts .. first from point B to the spring and secondly from touching the spring to full compression.
Initially I calculated speed on reaching spring using ..
v(sqrd) = v(initial)(sqrd) + (2a) x (distance) (a calculated from friction f=ma).
Next I noted that since we have a speed value there must be an initial KE value too. Also there is now no gravitational PE value (no height) this is zero.
the other forces at work here comprise of kinetic friction alone.
So using the general energy equations here I got
K(1) + U(1) + W(friction) = K(2) + U(2)
Where K(1), K(2) are initial and final kinetic energies of stone
U(1), U(2) are initial and final potential energies of stone
W(friction) is work done by friction (f x distance (100 m))
U(1) and U(2) are zero ...
from this we get a value for final KE which I use as initial KE for next phase which is encountering the spring.
Now at the spring (100 m into the stone's travel) using the energy equation for elastic potential energy ..
K(1) + Uel(1) + (-W(friction)) = K(2) +Uel(2)
where K(1) here is just the K(2) value from previous calculations
K(2) = 0 (stone comes to rest)
Uel(1) = 0 (spring uncompressed)
Uel(2) = 1/2 k x(sqrd)
This gives .. 1/2 m v(sqrd) - [(0.2) mgx] = 1/2 (2.0) x(sqrd)
= ((0.5)(15.0)(10.4)(sqrd) - 29.4x = x(sqrd)
= 811.2 - 29.4x = x(sqrd)
NOTE - The friction term is a term in x as I am calculating the work done by friction over the unknown spring-compression distance.
The way I attempted to solve this is by rearranging the terms to give a quadratic equation .. x(sqrd) +29.4x - 811.2 = 0
then solve using the rule ... ax(sqrd) + bx+ c use
b+/- sqrt((b(sqrd) - 4ac)/2a)
However I get nonsense solutions when I do this.
I am at a loss as to where to go from here .. can anyone help please.