Answer:
1) 0.43 meters per second
2) 0.21 meters per second
3) 1.02 [tex] \frac{m}{s^{2}}[/tex]
4) 0.66 seconds
Explanation:
part 1
By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):
[tex]K=U [/tex]
[tex] \frac{mv^2}{2}=\frac{kx_{max}^2}{2}[/tex]
With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v
[tex]v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s} [/tex]
part 2
Again by conservation of energy we have kinetic energy equal potential energy:
[tex] \frac{mv^2}{2}=\frac{kx_{max}^2}{2}=[/tex]
[tex]v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s} [/tex]
part 3
Acceleration can be find using Newton's second law:
[tex]F=ma [/tex]
with F the force, m the mass and a the acceleration, but elastic force is -kx, so:
[tex]-kx=ma [/tex]
[tex] a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}[/tex]
part 4
The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:
[tex] T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s [/tex]
So between 0 and 4.5 cm we have half a period:
[tex]t=\frac{T}{2}=0.66s [/tex]
