Speed of box attached to spring

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The discussion focuses on calculating the speed of an 87-g box attached to a spring with a force constant of 82 N/m, initially compressed 11 cm and then released. The first part of the problem involves finding the speed of the box when the spring is stretched by 7.0 cm past the equilibrium, with energy conservation principles applied. The calculated speed at this point is approximately 0.55 m/s, but there are suggestions to refine the approach using Newton's second law and integrating acceleration over the motion. The maximum speed of the box is also a point of interest, with energy conservation being emphasized as the correct method for solving the problem. The conversation highlights the importance of maintaining symbolic solutions before substituting numerical values to avoid errors.
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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)

Homework Equations


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
 
Last edited:
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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 don't 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).
 
TalibanNinja said:
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.
 
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