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?

6.5 m/s^2
2.7 m/s^2
26 m/s^2
13 m/s^2

I think it's A, but I'm unsure.

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?

6.5 m/s^2
2.7 m/s^2
26 m/s^2
13 m/s^2

I think it's A, but I'm unsure.

shamim · Answer

U know 
F=kx=?

shamim · Answer

Given k nd x

shamim · Answer

Then use F=ma
a=?

kaptain_mittens · Answer

So do F=kx as shamim said and you end up with:  F=.25(50), then after getting F  you do F=1kg(a)

anonymous · Answer

Okay @kaptain_mittens so I get
f=12.5
12.5=1kg(a)?

shamim · Answer

Tell us
a=?

anonymous · Answer

I think that's why I'm confused :( @shamim

anonymous · Answer

The force of a spring can be given as:
$$\huge \text{F}_\text{spring}=k \Delta x$$Given your scenario, we have$$\huge \text{F}_\text{spring}=(0.25)(50)=12.5$$However, given that it's a force, any force will follow Newton's 2nd Law, which is$$\huge \text{F}=ma$$Using our spring force that we calculated earlier, we can see that$$\huge 12.5=(1)(a) \rightarrow a=12.5 \frac{\text{m}}{\text{s}^2}$$