# Elastic potential energy of toy car

## 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.

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Method 2: (incorrect method)
F=kx
=50 x 0.1
= 5N
F=ma <---(THIS force is variable with displacement which is assumed to be const)
=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.
F=-kx, as spring returns into original position,x decreases so F also decreases,but in the above its assumes const.

Fermat
Homework Helper
Method 2 is wrong because F, the spring force is not a constant value as the spring compresses/expands.
The maximum spring force, after compression of an amount x, is F1 = kx.The minimum spring force, when the spring has expanded back to its natural length, is F2 = 0. Since the spring force F is a linear function of the compression, then the average force is just the simple average of the max and min vaues.
So Fav = 0.5(F1 + F2)
Fav = 0.5(kx + 0)
Fav = 0.5kx

Work done is the average force times the distance moved by the average force,

W = Fav*x
W = 0.5kx²