Question

1.60 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 13.0 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

Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2$

$A^2=\dfrac{mv^2}{k}$

Put the value into the relation

$A^2=\dfrac{1.60\times13.0^2}{315}$

$A=\sqrt{0.858}$

$A=0.926\ m$

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

$a=A\omega^2$

$a=A\times\dfrac{k}{m}$

Put the value into the formula

$a=0.926\times\dfrac{315}{1.60}$

$a=182.31\ m/s^2$

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

$F=ma$

Put the value into the formula

$F= 1.60\times182.31$

$F=291.69\ N$

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.