Recent content by fonzi

DISREGARD THE LAST POST ITS WRONG. this is a solved problem thanks for the help everyone.

Problem 8 is for the Mass 3m find the time that is required to reach the floor (from the top of the ramp to the floor). I use three formulas v0sin theta =V0y, y = v0yt =1/2 at^2, and the quadratic formula and you solve for t. Problem 9 is Calculate the velocity of the mass 3m at the point of...

thanks tiny tim problem 1 is the same as before for problem two and three we use the formula PE = (mV1^2)/2 + (3mV2^2)/2 to calculate conservation of energy and mv1 =3mv2 to calculate conservation of momentum. this will give you 8.39 m/s for problem 2 and 2.8 m/s for problem three. problem 4...

I'm running out of time can't you just tell me what I need to do? I can do the rest and I have shown an attempt at the problem. Also I know that 1,4,5,and 7 are correct. and their were actually 11 problems and I can solve those If I can find the answer to 2 and 3.

I'm sorry I can't figure this out.

Can you give me a formula or a procedure that I can use, cause I know I'm not doing this right, but I want to know the right way.

OK I'm not quite sure how to do that, But this is my attempt. Because of conservation of momentum I know PE = (.5V1^2)/2 + (1.5V2^2)/2 but that leaves me with two variables, so what do I do? M =.5 and 3M = 1.5, which means I am using the formula KE = (mv^2)/2

why doesn't my answer take into account the conservation of energy and momentum/how do It the right way?

final answers: for problem 1: PE =KE, where PE = potential energy and KE = kinetic energy formula two equals .5K delta X^2. this means PE = (750*.25^2)/2 = 23.44 J for problem 2 I believe you use V = square root (2KE)/m = square root 2(23.44)/.5 = 9.68m/s I believe problem 3 is solved the same...

so T should = .43 because y=.5at^2, so t = square root y/(.5a)

The way I see it I found the x direction velocity in problem 3 so I need to find V that accommodates two dimensions (the x as well as the y). so i am using the formula v0cos 20 = vx so vx/cos20 = v0 which gives me 5.55m/s. So I'm wondering if my method is correct?

I've already solved that, I'm on step 6 but the answer is 22.44, all the info you need is in the first, second, and fourth post.

I'm confused how does that help me find the speed of mass 3m when it leaves the ramp? Your answer seems to be the perspective I need to look at this problem, is their something wrong with how I have been solving the problems? If so where did I mess up and if not using work/energy formulas how do...

I believe I made a mistake on # 5 so what I think I was suppose to do was take the work of the spring - the work of friction = (mv^2)/2 now multiply the new work by 2 divide by m and square root both sides and you get velocity. so the answer should be 9.568 round to 9.57.

Is anyone there?