Compressed spring with frictional forces problem

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SUMMARY

The problem involves a 2.50 kg book compressed against a spring with a force constant of 250 N/m, compressed by 0.250 m, and sliding on a surface with a kinetic friction coefficient of 0.30. The work-energy theorem is applied to determine the distance the book travels before stopping. The initial potential energy stored in the spring is calculated as 1/2 * k * x^2, while the frictional force is defined as F_R = μN = μmg. The solution requires equating the initial energy to the work done against friction to find the stopping distance.

PREREQUISITES
  • Understanding of the work-energy theorem
  • Knowledge of spring potential energy calculations
  • Familiarity with kinetic friction and its equations
  • Basic application of Newton's laws of motion
NEXT STEPS
  • Calculate the initial potential energy of the spring using the formula U(s) = 1/2 k x^2
  • Determine the frictional force using F_R = μN and analyze its impact on motion
  • Apply the work-energy principle to find the distance traveled before stopping
  • Explore similar problems involving springs and friction for practice
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of work-energy applications in real-world scenarios.

offbeatjumi
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Homework Statement



A book (mass 2.50 kg) is forced against horizontal spring (negligible mass) with force constant 250 N/m, the spring is compressed 0.250m. When it's released the textbook slides horizontally across a surface with kinetic friction coeff = 0.30. How far does the book move from its original position before stopping?

Homework Equations



W = 1/2kX^2... F(s) = kx. U(s) = 1/2kx^2

The Attempt at a Solution



I know I need to use the work energy theorem here.
I was trying to use E2 + W(non-cons) = E1 since there is a nonconservative force in this problem. But isn't the initial and final kinetic energy both zero? I tried using Newton's 2nd law as well... F(net,x) = F(s) - f(k). But at the time when its released, it is not static equillibrium and I can't solve that equation. Unless... does f(k) = (mu)k*n (normal force?) where n = mg.
If anyone can help me set a foot in the correct direction I'd really appreciate it. Thanks!
 
Last edited:
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offbeatjumi said:

The Attempt at a Solution



I know I need to use the work energy theorem here.
I was trying to use E2 + W(non-cons) = E1 since there is a nonconservative force in this problem. But isn't the initial and final kinetic energy both zero? I tried using Newton's 2nd law as well... F(net,x) = F(s) - f(k). But at the time when its released, it is not static equillibrium and I can't solve that equation. Unless... does f(k) = (mu)k*n (normal force?) where n = mg.
If anyone can help me set a foot in the correct direction I'd really appreciate it. Thanks!

Right good, this is correct. If frictional force FR= μN = μmg, and the initial energy is 1/2kx2, what is this converted into during the motion in general?
 

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