Question

A 1.0 kg weight suspended from a spring is pulled to 0.25 m below its equilibrium point.

If the spring has a spring constant (k) of 50.0 N/m, at what rate will the mass accelerate when it is released?
O 200 m/s²
O 12.5 m/s²
○ 2.7 m/s²
O 6.5 m/s²

Answer 1

B.12.5m/s²

Step-by-step explanation:

Greetings!!!

Given values :-

weight= 1.0kg

Extension (x)= 0.25m

spring constant (k)= 50.0N/m

Required value :-

Acceleration= ?

Solution:-

`firstly.recall \: the \: hooks \: law`

• F=Kx

`substitute \: known \: varabiles \: into \: the \: equation`

• F= (50)(0.25)

`solve \: for \: force`

• F= 12.5N

`secondly \: reall \: the \: newtons \: second \: law`

• F= ma

`substitute \: known \: variables \: into \: the \: equation \:`

• (12.5)= (1.0)a

`solve \: for \: acceleration \:`

• a= 12.m/s²

Hope it helps!!!