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

(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.

Step-by-step 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

`(1)/(2)mv^2=(1)/(2)kA^2`

`A^2=(mv^2)/(k)`

Put the value into the relation

`A^2=(1.60*13.0^2)/(315)`

`A=√(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*(k)/(m)`

Put the value into the formula

`a=0.926*(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*182.31`

`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.