Recent content by swannyboy

Now if I try using momentum.../ F(delta_t) = m (delta_v) = (10kg)(4.9m/s) Impulse = 49 kgm/s 49kgm/s = F (0.125s) F = 392 N Fnet = fg + fa Fnet = 49N + 392N Fnet = 441N So, is that what the tension would be?

Homework Statement A crate is hanging from a cable as shown in the diagram. A monkey that is inside the crate, sits on a ledge that is 1.225m above the bottom of the crate. The monkey steps off the ledge and when he hits the bottom of the ledge, it takes .125 s to stop. The tension in the...

I think it must be an equivalent form. I got that it was the square root of 1/3. I've had this confirmed by 2 people now. Edit: Nope, you're right. Using exact values, it still works out to be the square root of 1/3. Well... I tried it using the form you got. I got slightly more than the...

Okay, if you want to find speed as a function of x, then you want to rewrite the equation in the form of v^2 =... So when I do that, I get V^2 = (50Nmx^2 -10Nx)/1.5kg When I did this before, I forgot the 1.5 kilograms, so I went an entirely different direction. Now when I do this...

so mgx = mgx = 1/2(m1+m2) V^2 + 1/2kx^2 + Ffx I don't see why I can't figure it out from here. mgx = KE + 1/2kx^2 + Ffx -KE = 1/2kx^2 + Ffx -mgx -KE = 25Nmx^2 + 4.8Nx -9.8Nx KE = 5.0Nx -25x^2 KE =5x(1-5x) Yes, I know I currently don't have units.

1/2 (m1 + m2) v^2

So if m1 = 1kg m2 = .5kg m1gx = 1/2 m1 v^2 + 1/2 m2 V^2 + 1/2kx^2 + Ff xor maybe m1gx = 1/2 m1m2 v^2 + 1/2kx^2 + Ffx Or m1gx = m1/2m2 v^2 + 1/2kx^2 + Ffx

If it starts at rest, the value of the constant is zero. Also, with the 1/2(m?)V^2, I think this mass would be the 0.5kg mass. I'm just guessing because the mass used for mgx must be the one that has the gravitational force acting on it, and the other one must be used some time, right?

mgh + 1/2mv^2 + 1/2kx^2 + F(friction)x = constant

So here are all the equations for where energy can be found: PE(s)=1/2kx^2 KE = 1/2mv^2 PE(g) = mgh So I guess the expression for friction would be 4.8N x distance in meters. It's energy lost to friction, so it would be negative, right?

I mentioned another place energy can go? I guess I got these: kinetic energy of the moving parts gravitational potential energy potential energy in the spring What else is there? Energy of friction? rotational energy in the pulley? I've looked through my posts, and I don't see where I've...

I think part of it may be because I'm tired right now, and I'm freaking out a little bit because I have a test on Monday, and it's 15 minutes 'til Sunday, but I really don't see how to do that right now. I mean I can try. If I do this at x= -0.01m when the start = 0, then: PE(spring) = 1/2kx^2...

Okay, so here's what I've done since. I'm still not getting the answer, but I've set the two accelerations equal to each other. a(1kg) = -a(.5kg) (9.8N - T)/1kg = (-T)/.5kg 1.5T = 4.9N T = 3.26 N Fnet = fg - T - f(friction) Fnet = 9.8N - 3.26N - 4.8N Fnet = 1.74 N F = ma 1.74N = 1kga a = 1.74...

Yes, that's right. x would be the number of meters.

Homework Statement http://img524.imageshack.us/i/physics123.png/" The image is here. Sorry that I couldn't attach a file. The file I made was too big. In the diagram given, the pulley is frictionless and the force of friction between the .5kg mass and the table is 4.8 N. The force...