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ziro4ka [17]
3 years ago
13

A 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equili

brium position and released, what is its maximum speed while it oscillates?
Physics
1 answer:
PolarNik [594]3 years ago
3 0
The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
E=U_{max}= \frac{1}{2}k(x_{max})^2

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
E=K_{max}= \frac{1}{2}m(v_{max})^2

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
U_{max}=K_{max}
\frac{1}{2}k(x_{max})^2 =  \frac{1}{2}m(v_{max})^2
v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s
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