# Speed of box attached to spring

## Homework Statement

An 87-g box is attached to a spring with a force constant of 82 N/m. The spring is compressed 11 cm from the equilibrium position and the system is released.
(a) What is the speed of the box when the spring is stretched by 7.0 cm passing the equilibrium?
(b) What is the maximum speed of the box? (A, 10 marks)

Ek=1/2mv^2
Ee=1/2kx^2

## The Attempt at a Solution

a)
Ek=Ee
1/2[(82N/m)(0.007m - (-0.011m))^2
=0.013284 J
0.013284J=1/2mv^2
√[(0.013284J)(2)]/(0.087kg)=v
v=0.55m/s

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Consider the forces, and use newtons 2nd law:
kx = ma
Integrate the acceleration once from -11 to 7(can u see why?)

i have found the acceleration using your suggestion but now i dont know how to find the speed of the box. the acceleration that is calculated was when the spring had recoiled and pushed the box out.
F=kx
(82N/m)(0.007M)/(0.087kg)=a
a=6.597 m/s^2

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Having found the acceleration you can integrate over it. And never insert numbers until you have a symbolic solution, because here you took some information out by doing so.

you see x causes the velocity to go like a parabola, while your approach is good for constant acceleration(velocity straight line).

Doc Al
Mentor
a)
Ek=Ee
Energy conservation says: Ek + Ee = constant
Thus: Ek1 + Ee1 = Ek2 + Ee2
1/2[(82N/m)(0.007m - (-0.011m))^2
=0.013284 J
Several errors here. Redo it. But energy conservation is the right idea.