Calculating Spring Compression and Velocity with Friction

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Homework Help Overview

The problem involves a block on a horizontal surface interacting with a spring, focusing on calculating the compression of the spring and the velocity of the block as it separates from the spring. The context includes frictional forces affecting the energy transfer during the spring's release.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the work-energy principle and the relationship between potential energy stored in the spring and kinetic energy of the block. Questions arise regarding the impact of friction on the calculations.

Discussion Status

Some participants have attempted calculations based on the spring's potential energy but have identified discrepancies due to the presence of friction. Guidance has been offered to consider the work done by friction in the analysis.

Contextual Notes

Participants are navigating the implications of friction on the energy calculations, with specific values provided for the coefficient of kinetic friction and the spring constant. There is an acknowledgment that the initial approach assumed a frictionless scenario, which is not applicable here.

Markrosoft
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having trouble with this problem... any ideas?

A 2.798-kg block is on a horizontal surface with muk = 0.170, and is in contact with a lightweight spring with a spring constant of 775 N/m which is compressed. Upon release, the spring does 0.969 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed. What is the velocity of the mass as it breaks contact with the spring?

i need this equation: 1/2 kx^2 ... but I'm not sure where to go from here...
 
Last edited:
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What are some universal equations for work you could use here? Think about how energy is transferred from potential to kinetic when the spring is released and allowed to return to equilibrium.
 
i tried this:
1/2 k x^2=E ...0.969 J = 1/2 (775) (x)^2 ...x = 0.05m

but it's wrong... does anyone know what I'm doing wrong?
 
That would be correct if the surface were frictionless. Try to factor in the work done by friction. Remember that [itex]W=Fd\cos\theta[/itex] and [itex]F_{kinetic~friction}=\mu_{kinetic}F_{normal}[/itex].
 
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