Recent content by ewarsaw

Yeah. How did you get that? I'm so confused.

I think I got the answer, but I'm not sure if it's right. I found the force applied to the box, the force of the block on the table, and the acceleration of the block. FN=FA-μ(mg) FN=ma FA-μ(mg)=ma 9.81 N-μ(9.81 m/s2)(2.073 kg)=2.073 kg(3.799 m/s2) μ=.095 Do you think that's right?

I used conservation of momentum. m1v1=m2v2 1 kg(1.08 m/s)=2.073 kg(v) v=.523 m/s

Ok, I got that and I got .523 m/s for the velocity of the block after it's hit, but I don't know what to do after that.

Probably. How did you get that?

2.58 m/s ?

It turned into kinetic energy?

It transferred to the block.

Yeah, that's the situation. Sorry for not explaining.

Homework Statement Calculate the coefficient of friction between the wooden block and the table top, using a pendulum to hit the block. Mass of pendulum weight = 1 kg Mass of wooden block (with weight on top) = 2.073 kg Release height of pendulum = .06 m Pendulum's distance from...