# Speed of box attached to spring

• TalibanNinja
In summary, the problem involves an 87-g box attached to a spring with a force constant of 82 N/m. The spring is compressed 11 cm from equilibrium and the system is released. The question asks for the speed of the box when the spring is stretched by 7.0 cm past equilibrium, as well as the maximum speed of the box. Using energy conservation, the speed is calculated to be 0.55 m/s.
TalibanNinja

## 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

Last edited:
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

Last edited:
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.

## 1. What is the speed of a box attached to a spring?

The speed of a box attached to a spring depends on various factors such as the mass of the box, the stiffness of the spring, and the initial displacement of the box from its equilibrium position. It can be calculated using the equation v = √(k/m) x A, where k is the spring constant, m is the mass of the box, and A is the amplitude of the oscillation.

## 2. How does the spring constant affect the speed of the box?

The spring constant, denoted by k, determines the stiffness of the spring. A higher spring constant means a stiffer spring, which in turn leads to a higher speed of the box. This is because a stiffer spring can store more potential energy, which is converted into kinetic energy as the box oscillates.

## 3. Can the speed of the box attached to a spring be greater than the speed of sound?

No, the speed of the box attached to a spring cannot exceed the speed of sound. This is because the speed of sound is the maximum speed at which energy can be transmitted through a medium, and the box-spring system is still limited by the properties of the spring and the box.

## 4. How does the mass of the box affect the speed of the system?

The mass of the box has a direct impact on the speed of the system. As the mass increases, the speed decreases, and vice versa. This is because a heavier box requires more energy to move, which means the spring will have to exert more force to achieve the same amplitude of oscillation.

## 5. Can the speed of the box attached to a spring be constant?

No, the speed of the box attached to a spring is not constant. It varies as the box oscillates back and forth, with the maximum speed occurring at the equilibrium position (where the spring is neither compressed nor extended) and decreasing as the box moves away from this position. However, the average speed of the box over one complete oscillation remains constant.

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