- #1
Destructo_Dav
- 3
- 0
So I checked out the physics class I was taking on ratemyprofessor.com and John Ross was the guy who was teaching it. He had good reviews, so I figured I'll take his class. Well, a TA is teaching it, and I won't name the person, but he isn't really a teacher--even with his PhD in astrophysics. Needless to say, I'll probably be needing quite a bit of help. Thanks in advance for anyone who does help.
A 12g bullet is fired in a 87.8g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 9.8m before coming to rest. The acceleration of gravity is 9.8 m/s^2
If the coefficient of friction between the surface and the block is .5, find the speed of the bullet before impact.
Answer: 81.5033 m/s
Normal Force = mg
Work = Distance * Force
Another one with velocity in the variable
The farthest I got was finding the frictional force that opposes the motion of the block.
Normal Force = (.012+.0878)(9.8) = .97804 in the upward direction
Friction Force = (.5)(.97804) = .48902
I suppose I could turn that into work, which would be:
Work = (.48902)(9.8) = 4.79240 in the westward direction
That gives me the force that opposes the motion of the block, but I don't think that gets me anywhere closer to the velocity. This is where I am stuck, and although I know the answer, I will have to know how to do this for the test. Any guidance will be much appreciated.
Homework Statement
A 12g bullet is fired in a 87.8g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 9.8m before coming to rest. The acceleration of gravity is 9.8 m/s^2
If the coefficient of friction between the surface and the block is .5, find the speed of the bullet before impact.
Answer: 81.5033 m/s
Homework Equations
Normal Force = mg
Work = Distance * Force
Another one with velocity in the variable
The Attempt at a Solution
The farthest I got was finding the frictional force that opposes the motion of the block.
Normal Force = (.012+.0878)(9.8) = .97804 in the upward direction
Friction Force = (.5)(.97804) = .48902
I suppose I could turn that into work, which would be:
Work = (.48902)(9.8) = 4.79240 in the westward direction
That gives me the force that opposes the motion of the block, but I don't think that gets me anywhere closer to the velocity. This is where I am stuck, and although I know the answer, I will have to know how to do this for the test. Any guidance will be much appreciated.
