A 11.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leavin
ID: 1348065 • Letter: A
Question
A 11.0 kg 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 Aand 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 2.30 N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.
NEED PART B:
How far will the stone compress the spring?
ANSWER: ______________ m
Explanation / Answer
Initial KE = m v^2 /2 = 11 x 11^2 /2 = 665.5 J
work done by gravity = mgh = 11 x 9.81 x 20 = 2158.2 J
work done friction = uk mg d = 0.20 x 11 x 9.81 x (100 +d)= - (2158.2 + 21.58d )
work done by sprijng = - kd^2 /2
using work energy theorem, ,
2158.2 - (2158.2 + 21.58d ) - kd^2 /2 = 0 - 665.5
1.15d^2 + 21.58d - 665.5 = 0
d =16.44 m
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