2. A vertical spring has one end attached to the ceiling and a 2.50-kg weight at
ID: 1395269 • Letter: 2
Question
2. A vertical spring has one end attached to the ceiling and a 2.50-kg weight attached to the other one. When the system is at rest, the spring is stretched by 20.0 cm. Now let the weight drop from a position in which the spring is not deformed at all. Use the conservation of energy law to find: a) how fast the weight is moving after it drops 30.0 cm; b) How far down the weight will drop before starting to come back. What is the weight?s acceleration when it's at the lowest position (give the magnitude and show the direction)?Explanation / Answer
a) Let K is the spring constant.
In the equilbrium
Fg = F_spring
m*g = k*x
==> k = m*g/x
= 2.5*9.8/0.2
= 122.5 m/s
Apply conservation of energy
m*g*h = 0.5*k*h^2 + 0.5*m*v^2
2.5*9.8*0.3 = 0.5*122.5*0.3^2 + 0.5*2.5*v^2
7.35 = 5.51 + 1.25*v^2
v = sqrt((7.35-5.51)/1.25)
= 1.21 m/s <<<<<<<--------Answer
b) 0.5*k*h^2 = m*g*h
h = 2*m*g/k
= 2*2.5*9.8/122.5
= 0.4 m or 40 cm <<<<<<<--------Answer
c) Fnet = k*x - m*g
m*a = k*x - m*g
a = k*x/m - g
= 122.5*0.4/2.5 - 9.8
= 9.8 m/s^2 (upward) <<<<<<<--------Answer
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