- #1
Trojanof01
- 13
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Bullet is fired at a block resting on a table. The velocity of the bullet is 390 m/s and has a mass of .007kg. The mass of the block is .850kg. The bullets penetrates the block and exits it with a velocity of 130 m/s causing the block to slide a distance of .410 m.
What is the coefficient of friction?
So, I thought maybe I could use the equation F= (mu)n -------> F = (mu)mg
and v^2 = (Vo)^2 + 2a(x-xo);
I plugged in the final and initial velocities along with the displacement and solved for a.
Plugged a into the equation F=ma to get the force then used that number and set it equal to (mu)mg to solve for (mu). Ended up with some number that was way too large to be a coeff. of friction.
I'm guessing it was a mistake to assume the velocity of the block at the split second after the bullet exited, was the same as the velocity of the bullet.
Any help would be appreciated...
What is the coefficient of friction?
So, I thought maybe I could use the equation F= (mu)n -------> F = (mu)mg
and v^2 = (Vo)^2 + 2a(x-xo);
I plugged in the final and initial velocities along with the displacement and solved for a.
Plugged a into the equation F=ma to get the force then used that number and set it equal to (mu)mg to solve for (mu). Ended up with some number that was way too large to be a coeff. of friction.
I'm guessing it was a mistake to assume the velocity of the block at the split second after the bullet exited, was the same as the velocity of the bullet.
Any help would be appreciated...