What is the speed of the block after the collision?

AI Thread Summary
The discussion revolves around calculating the speed of a block after a bullet embeds itself in it during a collision. The block travels 1.70 m post-collision, with the bullet weighing 5.26 g and the block 2.40 kg, while the coefficient of kinetic friction is 0.30. The initial attempt to solve the problem incorrectly equates kinetic energy with friction force, leading to confusion about the distance traveled. A correct approach involves using the work-energy principle to relate the work done by friction to the kinetic energy of the block. Understanding this relationship will help determine the block's speed immediately after the collision.
huynhtn2
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Homework Statement


When a bullet strikes a block of wood originally at rest, the bullet becomes embedded into the wood block which travels 1.70 m after the collision. If the mass of the bullet is 5.26 g and the mass of the block is 2.40 kg, what is the speed of the block immediately after the collision? The coefficient of kinetic friction between the block and the surface is 0.30.



Homework Equations


ke=0.5mv^2
Ffr= ukmg


The Attempt at a Solution


Ke = Ffriction
0.5mv^2= uk mg
v= square root (2(9.81)(0.3))
v= 2.43 m/s

i don't know what to do with the 1.70 m distance
 
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huynhtn2 said:

Homework Statement


When a bullet strikes a block of wood originally at rest, the bullet becomes embedded into the wood block which travels 1.70 m after the collision. If the mass of the bullet is 5.26 g and the mass of the block is 2.40 kg, what is the speed of the block immediately after the collision? The coefficient of kinetic friction between the block and the surface is 0.30.

Homework Equations


ke=0.5mv^2
Ffr= ukmg

The Attempt at a Solution


Ke = Ffriction
0.5mv^2= uk mg
v= square root (2(9.81)(0.3))
v= 2.43 m/s

i don't know what to do with the 1.70 m distance

Your second formula in the attempt at a solution is comparing energy and force. The units do not match. You can only compare an energy term with another energy term, or a force with another force.

How can you get an energy value using the friction force? Thinking about this will give you the solution.

Cheers -- sylas
 
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