Max velocity of blocks after spring pushes them apart

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SUMMARY

The discussion focuses on calculating the maximum velocities of two blocks (0.250 kg and 0.490 kg) after being pushed apart by a compressed spring with a force constant of 3.75 N/m. The calculations involve the spring's potential energy and the work done against friction, using the equation 1/2Kx^2 = 1/2mv^2 + fd. The maximum velocities were found to be 3.2 m/s for the 0.250 kg block and 0.23 m/s for the 0.490 kg block under specific coefficients of kinetic friction.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of kinetic friction and its coefficients
  • Familiarity with energy conservation principles in physics
  • Ability to solve quadratic equations
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  • Study the effects of varying coefficients of kinetic friction on block motion
  • Learn about energy conservation in elastic systems
  • Explore the relationship between mass, force, and acceleration using Newton's laws
  • Investigate real-world applications of springs in mechanical systems
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Homework Statement


A light spring of force constant 3.75 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.490 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is greater than the coefficient of kinetic friction. Let the positive direction point to the right.

(a) For the 0.250 kg block:
At µk = 0.000, the maximum velocity is _____ m/s
At µk = 0.100, the maximum velocity is _____ m/s
At µk = 0.454, the maximum velocity is _____ m/s

(b) For the 0.490 kg block:
At µk = 0.000, the maximum velocity is _____ m/s
At µk = 0.100, the maximum velocity is _____ m/s
At µk = 0.454, the maximum velocity is _____ m/s

Homework Equations



1/2Kx^2=1/2mv^2+fd

The Attempt at a Solution


For µk = 0.000 and m = 0.250 kg, I get:
1/2(3.75)(0.08^2)=1/2(0.25)v^2 + 0
solving for v i get: v=0.309 and v=-0.31
since the 0.250 kg block is moving to the left the answer should be -0.31 m/s, but the answer is incorrect.
 
Physics news on Phys.org
a)
velocity of mass 0.49 is 0.23 m/s
velocity of mass 0.25 is 3.2 m/s
 
how did you obtain those answers? I don't think they're correct.
 

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