Answer to Question #252519 in Physics for Jessy

Question

A 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

Given:

$m=2.00\:\rm kg$

$k=315\:\rm N/m$

$v_x(0)=-12.0\:\rm m/s$

The equation of motion

$$x(t)=-A\sin(\omega t)$$

$$\omega=\sqrt{k/m}=\sqrt{315/2.00}=12.5\:\rm rad/s$$

Hence

$$x(t)=-A\sin(12.5 t)$$

The velocity

$$v_x(t)=x'(t)=-12.5A\cos(12.5t)$$

(a) the amplitude of the motion

$$A=\frac{v_x(0)}{-12.5}=\frac{12.0}{12.5}=0.956\:\rm m$$

(b) the block's maximum acceleration

$$a_{x\max}=12.5^2A=12.5^2*0.956=149\:\rm m/s^2$$

(c) the maximum force the spring exerts on the block

$$F_{\max}=ma_{x\max}=149*2.00=298\:\rm N$$