Solving Kinetic Energy of Block After Bullet Passes Through

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SUMMARY

The discussion centers on calculating the kinetic energy of a wooden block after a bullet passes through it. A bullet with a mass of 6.00 g and an initial speed of 370 m/s exits the block at 120 m/s, while the block has a mass of 0.750 kg and slides 49.0 cm after the bullet's passage. The kinetic energy of the block immediately after the bullet exits is determined using the conservation of momentum and energy principles, resulting in a calculated kinetic energy of 14.06 J. The coefficient of kinetic friction between the block and the surface is established as 0.416, and the decrease in kinetic energy of the bullet is calculated to be 368 J.

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DPatel304
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Stuck on this one problem which states:
A bullet with a mass of 6.00 g, traveling horizontally with a speed of 370 m/s, is fired into a wooden block with mass of 0.750 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 m/s. The block slides a distance of 49.0 cm along the surface from its initial position.
Use g = 9.81 m/s^2 for the acceleration due to gravity.And I am trying to find this:
What is the kinetic energy of the block at the instant after the bullet passes through it?

If more information is needed, I can post it. Thanks in advance for your help.

Btw, the first portion of the question asked for the coefficient of friction, which I was able to get:
What is the coefficient of kinetic friction between block and surface?
0.416

and this:
What is the decrease in kinetic energy of the bullet?
368 J
 
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Can you find an expression for conservation of energy of the system?

Of course we are ignoring any deformation of the block or bullet.
 

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