Question

A wooden ball weighing 8.50 N is pushed into the air by a vertical spring with a spring constant of 1190 N/m . If the spring was compressed 0.170 m before pushing the ball straight up, what is the maximum height that the ball will reach above the initial position?

Answer 1

Using energy conservation:

KEi + PEi = KEf + PEf

KEi = 0, since initially spring was compressed and ball was at rest

KEf = 0, since at max height speed of ball will be zero,

PEi = (1/2)*k*x^2 = spring potential energy (assuming this as a reference point where GPE is zero)

PEf = m*g*(h + x) = Gravitational PE at 'h' height

So,

0 + (1/2)*k*x^2 = m*g*(h + x)

h = ((1/2)*k*x^2 - m*g*x)/(m*g)

m*g = Weight of ball = 8.50 N

k = 1190 N/m

So,

h = (0.5*1190*0.170^2 - 8.50*0.170)/8.50

h = 1.853 m = max height of ball

h + x = max height above initial position = 1.853 + 0.170 = 2.023 m = 2.0 m

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