Recent content by soccer_09

I figured it out. They give the weight as 17.0 N which is mg so I was doing g squared at first. When I got my answer with that, I added it to the initial distance and got it right. Got it on the 10th try out of 10 as well :) Thanks for trying though hehe. Was just a pure coincidence that the cos...

I already tried that and it was wrong. I got .8026 for my answer and I even tried doing sin and cos of 40 degrees. None of them worked. I looked on cramster and found the same problem since no one had been answering my other post and I found that they got the right answer using the formula I...

Homework Statement Block on Incline In Fig. 16-35, a block weighing 17.0 N is able to slide without friction on a 32.0° incline. It is connected to the top of the incline by a massless spring of unstretched length 0.475 m and spring constant 110 N/m. Figure 16-35 (a) How far down the...

Can someone please explain to me what I'm doing wrong with this problem? In Fig. 16-35, a block weighing 17.0 N is able to slide without friction on a 32.0° incline. It is connected to the top of the incline by a massless spring of unstretched length 0.475 m and spring constant 110 N/m...

Thanks :) I integrated from 0 to 4 of F(t) and got 10.4. using the constants I and R, I was able to get the correct answer. Thanks again.

Yep, I had a friend tell me I converted incorrectly. Thanks :) From there though, my alpha = 680 rad /s^2 The equation for rotational veloctiy is: omega(t) = omega (initial) + alpha (t) so with alpha = 680 rad/s^2 and t = 4s that should give me a rotational speed of 2720 rad/s...

Homework Statement A pulley, with a rotational inertia of 2.0 10-3 kg·m2 about its axle and a radius of 20 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t2, where F is in Newtons and t in seconds. The pulley is initially at...