Recent content by jnthn205

well thanks anyways...

Please doc don't give up on me now!

Or is the new force when friction is involved = 1500N/m-(1500N/m*.475)=787.5 N/m then use that in the equation? Like this: (1/2)(.15^2)/(2*9.8*sin(37))*787.5 Is this correct?

Awesome thanks! And to calculate the same thing, only this time with friction. I just plug in friction = .475 into the bottom of the equation from post 5? so (1/2)(.15^2)/(2*9.8*sin(37))*1500*.475 that doesn't seem right, the distance would be greater than it is with no friction. Or do I have...

173-45 =128? I only have 1 attempt left and I want to make sure this is correct before i input the answer

i found 143+30 so 183 so 183-45? so 138cm?

so the block ends up 143+45cm down the ramp? ugh I am so confused. Once you tell me I will know what your talking about from what you are saying you want me to subtract 143 from 45, but I did that and you said it was wrong...so I am lost.

I don't know, I thought that's what I was solving for. How do I solve for L then?

or no, would you subtract the total spring length 45 subtracted byhow much it compressed 15? So 45-15 = 30. So the block starts at 30cm. So add 30cm to 143cm?

the starting point is when the spring is fully compressed right? So it starts at 45cm? Then it goes 143 more cm? So do I add them? 45+143?

143-45?? so 98cm?

i need help fast! This is due in 2 hours

ok so (.15m)^2 / [2(2kg)(9.8m/s^2)(sin37)] 1500 N/m = 1.43m. But it wants the answer in cm, so 1.43 m = 143cm which was incorrect

ok so i have the formula for distance = x^2 / [2(mg sin(theta)] L which would be (15cm)^2 / [2(2000g)(980cm/s^2)(sin37)] 15 N/cm = .00143 This answer was wrong, but I think the formula is correct. I think I have my units messed up. Can someone please input the correct units?

Homework Statement A(n) 2000g block is pushed by an external force against a spring (with a 15N/cm spring constant) until the spring is compressed by 15cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 37degrees. Note: The spring lies along...