Finding the Coefficient of Kinetic Friction

[besenji]

This problem has been giving me some trouble, and I was wondering if anyone could help?

A 0.700 kg wood block is firmly attached to a very light horizontal spring (k = 200 N/m) as shown in the figure (Intro 1 figure) . This block-spring system, when compressed 4.7 cm and released, stretches out 2.5 cm beyond the equilibrium position before stopping and turning back.

What is the coefficient of kinetic friction between the block and the table?

I just can't seem to find the correct equation.

I know that it involves 1/2kx^2. I am just not sure how to find the uk using that.

Any bit of help is greatly appreciated.

 

[alphysicist]

Hi besenji,

This problem has been giving me some trouble, and I was wondering if anyone could help?

A 0.700 kg wood block is firmly attached to a very light horizontal spring (k = 200 N/m) as shown in the figure (Intro 1 figure) . This block-spring system, when compressed 4.7 cm and released, stretches out 2.5 cm beyond the equilibrium position before stopping and turning back.

What is the coefficient of kinetic friction between the block and the table?

I just can't seem to find the correct equation.

I know that it involves 1/2kx^2. I am just not sure how to find the uk using that.

Any bit of help is greatly appreciated.

I would suggest using the equation that relates non-conservative work to the change in energy. What do you get?

 

[thomate1]

I would first like to commend you for seeking help in solving this problem. It is important to recognize when we need assistance in understanding complex concepts or equations.

To find the coefficient of kinetic friction in this scenario, we can use the equation: μk = (kx^2)/(mg), where μk is the coefficient of kinetic friction, k is the spring constant, x is the distance the spring is compressed, m is the mass of the block, and g is the acceleration due to gravity.

In this problem, we know all the variables except for the coefficient of kinetic friction. We can substitute in the given values and solve for μk:

μk = (200 N/m * (4.7 cm / 100 cm)^2) / (0.700 kg * 9.8 m/s^2)

μk = 0.18

Therefore, the coefficient of kinetic friction between the block and the table is 0.18.

I hope this helps in your understanding of the problem. If you have any further questions, do not hesitate to ask for clarification. Keep up the good work in your scientific pursuits!