Asked by tom
A 1100 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 40 cm. What is the spring constant of the spring?
I don't understand what to do here
I don't understand what to do here
Answers
Answered by
Chanz
mgh = 1/2 kx^2
solve for k
solve for k
Answered by
Anonymous
1.1(9.8)(2.0)
=21.56
=21.56
Answered by
Jack
We know that potential energy is m*g*h (h = 2.0 + .4 = 2.4m)
and Hooke's formula is 1/2*k*x^2
In this problem when the safe drops its potential energy is turned into kinetic, therefore m*g*h = 1/2*k*x^2
Solve for k:
(2*m*g*h) / x^2
and Hooke's formula is 1/2*k*x^2
In this problem when the safe drops its potential energy is turned into kinetic, therefore m*g*h = 1/2*k*x^2
Solve for k:
(2*m*g*h) / x^2
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