What Is the Frictional Force Between Block and Table After Bullet Impact?

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SUMMARY

The discussion focuses on calculating the frictional force between a block of wood and a tabletop after a bullet embeds itself in the block. The block has a mass of 2 kg, and the bullet weighs 5 g, traveling at a speed of 150 m/s. The speed of the block after the bullet impact is determined to be 0.374 m/s. To find the frictional force, participants are advised to apply the work-energy principle, equating the initial kinetic energy to the work done against friction as the block slides 270 cm before stopping.

PREREQUISITES
  • Understanding of momentum conservation principles
  • Familiarity with kinetic energy calculations
  • Knowledge of frictional force equations (Ff = μ * Fn)
  • Basic grasp of work-energy principles
NEXT STEPS
  • Calculate the frictional force using both kinetic energy of the bullet and the combined kinetic energy of the bullet and block.
  • Explore the implications of neglecting heat energy in collision scenarios.
  • Review the work-energy theorem and its applications in friction problems.
  • Investigate different coefficients of friction for various materials to understand their impact on force calculations.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to explain concepts of momentum, energy conservation, and friction in real-world scenarios.

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Homework Statement


2 kg block of wood rests on a long tabletop. A 5g bullet moving horizontally with a speed of 150m/s is shot into the block and sticks into it. The block then slides 270cm along the table and stops
A) find the speed of the block ( answer: 0.374 m/s)
B)*** find the frictional force between block and table ( answer: )


Homework Equations



Ff = mew * Fn
J=p
Ft = mv

The Attempt at a Solution



after trying to find the solution from relevant momentum formulas, i am still completely stumped as to how i would find the frictional force between table and block

much appreciated if you read this over
 
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Well, think about what happens. A bullet with a certain kinetic energy comes along and embeds itself in a block. The combined block and bullet slide across the table. Where does that initial kinetic energy go? If you are allowed to neglect the heat produced in the block during the collision, don't you suppose it goes into work to overcome friction as the block slides to a stop? If you are not allowed to neglect the heat generated, then you need the velocity of the block plus bullet right after the collision that gives you the kinetic energy of the pair at the start and then that energy disappears the same way as before. Think work = force times distance. That should get you started with the second part. However...

The first part makes use of momentum conservation. It would be a useful exercise for you to calculate the answer to the second part both ways (from the KE of the bullet and from the KE of the bullet plus block right after collision.) Do you find a difference?
 

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