- #1
Dealucis
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Homework Statement
Find the amount which the spring is compressed after the mass comes to rest. Assume a kinetic coefficient of friction of 0.11 and an initial velocity along the surface of v = 4m/s.
Given values :
mass = 2.2kg
k = 500 N/m
delta = 0.036m
theta = 23 degrees
The attempt at a solution
Alright.. here goes
My approach to this problem is to find the velocity at where the block touches the spring, and make it as the initial velocity.
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Using a free body diagram, I got the following values.
F = mg = 21.56N
N = mg cos theta = (21.56)cos23 = 19.846N
friction force = N.coefficient = (19.846)(0.11) = 2.183N
force down the plane = mg sin theta = (21.56)sin23 = 8.424N
Adding up the force in the x direction (parallel to the slope), F = 8.424 + (-2.183) = 6.241N
since F = ma,
6.241N = (2.2kg)a
therefore, a = 2.837m/s^2 [sorry, I'm not familiar with the forums. the unit is metre per seconds squared]
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Now, with the acceleration, the velocity at when it touches the spring can be calculated :
final velocity squared = initial velocity squared + 2(acceleration)(distance)
v^2 = u^2 + 2(a)(r)
v^2 = 4^2 + 2(2.837)(0.036)
v = squareroot(16 + 0.204)
v = 4.025m/s
therefore, the velocity at when it touches the spring is 4.025m/s
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Using the work-energy equation,
T1 + U = T2
as when the final velocity is when the block comes to rest, the final velocity is 0, therefore
(mv^2)/2 + U = 0
17.821 +U = 0
therefore U = 17.821J
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Using the value, the compression length can be found:
17.821 = 0.5(k)(final compression length squared - initial compression length squared)
17.821 = 0.5(500)(x^2 - 0)
17.821 = 250(x^2)
0.071283 = x^2
x = squareroot(0.071283)
x = 0.267m
that was my answer, but it's wrong. Much help would be appreciated on where i went wrong and how to solve the problem. Thanks.