1) Maximum speed: 3.0 m/s, when the spring is at equilibrium position
Explanation:
The law of conservation of energy tells that the total mechanical energy of the system, given by the sum of elastic potential energy (U) and kinetic energy (K), is constant:
[tex]E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const.[/tex]
where
k is the spring constant
x is the compression/stretch of the spring with respect to its equilibrium position
m is the mass of the block
v is the speed of the block
The maximum speed of the system occurs when the kinetic energy is maximum, and therefore when the elastic potential energy is zero: U=0, so we have:
[tex]U=\frac{1}{2}kx^2=0[/tex]
which means when x=0, so when the spring is at its equilibrium position. At this point of the motion, we also have
[tex]E=K_{max}=\frac{1}{2}mv_{max}^2[/tex] (1)
The total mechanical energy of the system is given by the problem:
[tex]E=11.5 J[/tex]
which corresponds to the potential energy of the system when the spring is at its maximum compression (because there the kinetic energy is zero). Solving (1) for v_max, we find:
[tex]v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(11.5 J)}{2.50 kg}}=3.0 m/s[/tex]
2) Maximum acceleration: [tex]96 m/s^2[/tex], when the spring is at its equilibrium position
Explanation:
The restoring force on the spring is given by Hook's law:
[tex]F=kx[/tex]
Using Newton's second law, we can rewrite the force as product between mass and acceleration:
[tex]ma=kx[/tex]
so we have
[tex]a=\frac{kx}{m}[/tex]
We infer that the greatest acceleration, [tex]a_{max}[/tex], will occur also when the displacement is maximum, [tex]x_{max}[/tex]:
[tex]a_{max}=\frac{kx_{max}}{m}[/tex]
The maximum displacement occurs when the elastic potential energy is maximum and the kinetic energy is zero, K=0, so we have:
[tex]E=\frac{1}{2}kx_{max}^2\\x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(11.5 J)}{2500 N/m}}=0.096 m[/tex]
So, the greatest acceleration is
[tex]a=\frac{(2500 N/m)(0.096 m)}{2.5 kg}=96 m/s^2[/tex]
