(c) With y= h= 5R, at Pwe find 2255(3.20 10 kg)(9.80 m/s )(0.12 m)0.19 JUmgR−==×=.(d) With y= R, at Qwe have (3.20 10 kg)(9.80 m/s )(0.12 m)0.038 JgR−=(e) With y= 2R, at the top of the loop, we find 22(3.20 10 kg)(9.80 m/s )(0.12 m)0.075 JgR−=(f) The new information ()vi≠0 is not involved in any of the preceding computations; the above results are unchanged. 8. We use Eq. 7-12 for Wgand Eq. 8-9 for U.(a) The displacement between the initial point and Qhas a vertical component of
This is the end of the preview. Sign up
access the rest of the document.