Answer to Question #252656 in Mechanics | Relativity for bal

Question

2.00-kg frictionless block is attached to an ideal spring with force constant 315 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude of the motion, (b) the block's maximum acceleration, and (c) the maximum force the spring exerts on the block.

Answer 1

(a) We can find the amplitude from the law of conservation of energy:

$$PE_s=KE_s,$$ $$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2,$$ $$A=\sqrt{\dfrac{mv^2}{k}}=\sqrt{\dfrac{2.0\ kg\cdot(-12.0\ \dfrac{m}{s})^2}{315\ \dfrac{N}{m}}}=0.96\ m.$$

(b) Let's first find the angular frequency:

$$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{315\ \dfrac{N}{m}}{2.0\ kg}}=12.5\ \dfrac{rad}{s}.$$

The position of the block can be found as follows:

$$x(t)=Acos(\omega t+\theta).$$

Let's first find the velocity of the block:

$$v(t)=\dfrac{dx(t)}{dt}=-\omega Asin(\omega t+\theta).$$

Then, we can find the acceleration of the block:

$$a(t)=\dfrac{dv(t)}{dt}=-\omega^2Acos(\omega t+\theta).$$

The block's acceleration will be naximum when $cos(\omega t+\theta)=\pm1$. Therefore, we can write:

$$a_{max}=-\omega^2A=-(12.5\ \dfrac{rad}{s})^2\cdot0.96\ m=-150\ \dfrac{m}{s^2}.$$

The sign minus means that the acceleration of the block is directed in the opposite direction from displacement of the block.

(c) We can find maximum force the spring exerts on the block from the Newton's Second Law of Motion:

$$F_{max}=ma_{max}=2.0\ kg\cdot(-150\ \dfrac{m}{s^2})=-300\ N.$$

The sign minus means that the force that he spring exerts on the block is directed in the opposite direction from displacement of the block.