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.
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.
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?
square root of(.79/(.5*65))
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
To calculate the potential energy stored in the spring, you can use the formula for spring potential energy, which is given by U = 0.5 * k * x^2, where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the spring is held in compression by a force of 10 N, and the spring constant is 20 N/m. We can find the displacement x using Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement from its equilibrium position. The equation for Hooke's Law is F = k * x.
Rearranging this equation, we have x = F / k. Substituting the known values: x = 10 N / 20 N/m = 0.5 m.
Now we can calculate the potential energy by substituting the values of k and x into the formula: U = 0.5 * (20 N/m) * (0.5 m)^2 = 0.5 * 20 N/m * 0.25 m^2 = 0.5 J.
Therefore, the potential energy stored in the spring is 0.5 Joules.