Asked by Kyle
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
Answered by
Anonymous
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?
Answered by
hussam
square root of(.79/(.5*65))
Answered by
Henry
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.
Answered by
Hermosa
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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