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## Homework Statement

A toy car of mass 0.5kg is pushed agains a spring so that it is compressed by 0.1m. The spring obeys Hooke's Law and has a spring constant of 50N/m. When the toy car is released, what will its speed be at the instant that the spring returns to its natural length? Assume that there is no friction within the sprin and no frictional force resisting the motion of the toy car.

## Homework Equations

Hooke's law:

F=kx

E(elastic potential) = 0.5kx^2

F=ma

W=0.5mv^2 - 0.5mu^2

v^2=u^2 + 2ax

## The Attempt at a Solution

I tried two methods, however only one gave me the correct answer. My question here is, why is this the case? Is there something i assumed that i shouldn't have?

Method 1: ( the correct one)

E(elastic potential)=0.5kx^2

=0.5 x 50 x 0.1^2

=.25J

W=0.5mv^2 - 0.5mu^2

0.25=0.5x0.5x v^2

v=1m/s (Right)

Method 2: (incorrect method)

F=kx

=50 x 0.1

= 5N

F=ma

=5/0.5

=10m/s^2

v^2=u^2 + 2ax

V^2=2 x 10 x 0.1

v= 1.41m/s (Wrong)

Why is this the case? Thankyou.