Box with mass attached to spring and board

In summary, the problem involves a wood block attached to a horizontal spring (k=160N/m), which is compressed 4.0cm and released, stretching out 0.9 cm beyond its equilibrium position. The question asks for the coefficient of friction, which can be found using the equation (coefficient of friction)(Normal Force)(d)=1/2k(xf-xi)^2. The direction of compression or extension does not matter as long as one is positive and one is negative. The approach used is correct.
  • #1
12boone
21
0

Homework Statement



A 0.580Kg wood block is firmly attached to a very light horizontal spring. It is noted that the block-spring system (k=160N/m), when compressed 4.0cm and released, stretches out 0.9 cm beyond the equilibrium position before stopping and turning back.

Homework Equations



Conservation of Mechanical Energy

KEf+PEf=KEi+PEi

and

1/2kx^2

The Attempt at a Solution



Ok, What I tried to do is solve for the work of friction using

(coefficient of friction)(Normal Force)(d)=1/2k(xf-xi)^2. However, I do not understand what the problem is asking as far as distances. I converted the distance to m but I still do not know what they mean by compressed for 4 cm. Does that mean they pushed it 4 cm forward. Let it go and it went past its resting point .9 cm. If so is .9 negative? or is 4 cm negative?
 
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  • #2
12boone said:
However, I do not understand what the problem is asking as far as distances. I converted the distance to m but I still do not know what they mean by compressed for 4 cm. Does that mean they pushed it 4 cm forward. Let it go and it went past its resting point .9 cm. If so is .9 negative? or is 4 cm negative?
That would be correct.

However, you have failed to state what the question is asking you to find.
 
  • #3
what I am looking for is the coefficient of friction.
 
  • #4
12boone said:
what I am looking for is the coefficient of friction.
Ahh right. Looking back at the question, is doesn't matter which (extension/compression) you choose as positive and which you choose as negative provided that one is positive and one is negative.

Your approach is correct by the way.
 

What is a box with mass attached to spring and board?

A box with mass attached to spring and board is a simple physics experiment that demonstrates the principles of springs and simple harmonic motion. The setup consists of a box or weight attached to a spring, which is then attached to a board or surface.

How does the mass of the box affect the motion of the spring?

The mass of the box attached to the spring affects the amplitude, or maximum displacement, of the spring's oscillations. A heavier box will cause the spring to stretch further, resulting in larger oscillations. The mass also affects the frequency of the oscillations, with a heavier box causing the spring to oscillate at a slower rate.

What is the relationship between the spring constant and the force exerted on the box?

The spring constant, also known as the stiffness of the spring, is directly proportional to the force exerted on the box. This means that as the spring constant increases, the force exerted on the box also increases. This relationship is described by Hooke's Law: F = -kx, where F is the force, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

What factors affect the period of oscillation for the box-spring-board system?

The period of oscillation, or the time it takes for the spring to complete one full cycle of motion, is affected by several factors. These include the mass of the box, the stiffness of the spring (spring constant), and the gravitational acceleration of the Earth. The period is also affected by any external forces acting on the system, such as air resistance or friction.

How can the energy of the box-spring-board system be calculated?

The total energy of the box-spring-board system is the sum of the potential energy stored in the spring and the kinetic energy of the moving box. The potential energy can be calculated using the equation PE = 1/2kx^2, where k is the spring constant and x is the displacement of the spring. The kinetic energy can be calculated using the equation KE = 1/2mv^2, where m is the mass of the box and v is its velocity. The total energy of the system is constant, meaning it remains the same throughout the motion of the box and spring.

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