Question
a force of 10 N holds an ideal spring with a 20-N/m spring constant in compression. The potential energy stored in the spring is..? 0.5J, 2.5J, 5.0J, 10J, or 200J?
POtential energy? Wouldn't it be 1/2 kx^2,
but x= F/k, so 1/2 kx^2= 1/2 k*F^2/k^2 and that can be reduced.
POtential energy? Wouldn't it be 1/2 kx^2,
but x= F/k, so 1/2 kx^2= 1/2 k*F^2/k^2 and that can be reduced.
Answers
A 0.5-kg block attached to an ideal spring with a spring constant of 65N/m oscillates on a
horizontal frictionless surface. The total mechanical energy is 0.79 J. What is the greatest extension
of the spring from its equilibrium length?
horizontal frictionless surface. The total mechanical energy is 0.79 J. What is the greatest extension
of the spring from its equilibrium length?
square root of(.79/(.5*65))
a. PE = 1/2 * kx^2.
or d = 10 * 1m/20N = 0.5 m.
PE = 0.5*F*d = 0.5 * 10 * 0.5 = 2.5 J.
or d = 10 * 1m/20N = 0.5 m.
PE = 0.5*F*d = 0.5 * 10 * 0.5 = 2.5 J.
F=10N
k=20N/m
x=?
PE=(1/2)(k)(x^2)
x=F/k=10/20=0.50
PE=1/2(20)(0.50)^2
=2.5J
k=20N/m
x=?
PE=(1/2)(k)(x^2)
x=F/k=10/20=0.50
PE=1/2(20)(0.50)^2
=2.5J
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