- #1
FlukeATX
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Homework Statement
The figure shows an 7.9 kg stone at rest on a spring. The spring is compressed 10 cm by the stone. The stone is then pushed down an additional 32 cm and released. To what maximum height (in cm) does the stone rise from that position?
Homework Equations
k = mg/d
v = sqrt((kx^2)/m)
h = ((1/2)(v^2))/(g)
The Attempt at a Solution
So this is what I've tried:
k = mg/d
k = (7.9)(9.8) / (.1)
k = 774.2 N/m
v = sqrt((kx^2)/m)
v = sqrt((774.2*.32^2)/(7.9))
v = sqrt(79.27808/7.9)
v = 3.1678
h = (1/2v^2) / (g)
h = (1/2)(10.0352) / (9.8)
h = .512m
h = 51.2cm
Does this seem correct? I only have one shot left on the wiley plus and don't want to risk it if it's wrong. Appreciate the help!