# Differences

This shows you the differences between two versions of the page.

 phy131studiof15:lectures:m1p1sol [2015/09/30 17:22]mdawber [Question 1 Solution] phy131studiof15:lectures:m1p1sol [2015/09/30 17:23] (current)mdawber [Question 1 Solution] Both sides previous revision Previous revision 2015/09/30 17:23 mdawber [Question 1 Solution] 2015/09/30 17:22 mdawber [Question 1 Solution] 2015/09/30 17:22 mdawber [Question 1 Solution] 2015/09/30 10:34 mdawber [Question 1 Solution] 2015/09/29 09:11 mdawber 2015/09/29 08:58 mdawber created 2015/09/30 17:23 mdawber [Question 1 Solution] 2015/09/30 17:22 mdawber [Question 1 Solution] 2015/09/30 17:22 mdawber [Question 1 Solution] 2015/09/30 10:34 mdawber [Question 1 Solution] 2015/09/29 09:11 mdawber 2015/09/29 08:58 mdawber created Line 25: Line 25: For parts C-G consider a case where $m_{1}=$4 kg and $m_{2}$=1 kg, $\mu_{s}$=0.2 and $\mu_{k}$=0.15. For parts C-G consider a case where $m_{1}=$4 kg and $m_{2}$=1 kg, $\mu_{s}$=0.2 and $\mu_{k}$=0.15. - C. (5 points) If  $\theta=30^{o}$ what is the velocity and displacement of $m_{1}$, 0.5\,s after the system has been released from rest? + C. (5 points) If  $\theta=30^{o}$ what is the velocity and displacement of $m_{1}$, 0.5s after the system has been released from rest? $a=\frac{4+1\sin 30^{o}-0.15\cos30^{o}}{4+1}\times9.8=8.57\mathrm{ ms^{-2}}$ $a=\frac{4+1\sin 30^{o}-0.15\cos30^{o}}{4+1}\times9.8=8.57\mathrm{ ms^{-2}}$ 