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 phy131studiof15:lectures:m2p1sol [2015/10/27 13:21]mdawber [Question 2 Solution] phy131studiof15:lectures:m2p1sol [2015/10/27 14:25] (current)mdawber [Question 2 Solution] Both sides previous revision Previous revision 2015/10/27 14:25 mdawber [Question 2 Solution] 2015/10/27 13:21 mdawber [Question 2 Solution] 2015/10/20 11:46 mdawber 2015/10/20 11:46 mdawber created 2015/10/27 14:25 mdawber [Question 2 Solution] 2015/10/27 13:21 mdawber [Question 2 Solution] 2015/10/20 11:46 mdawber 2015/10/20 11:46 mdawber created Line 106: Line 106: F. (10 points) How much work does the person turning the wheel have to do to raise the bat by 1.5m? (Don't forget to include kinetic energy, the bat is moving with the velocity you found in part C when it has been raised by 1.5m) F. (10 points) How much work does the person turning the wheel have to do to raise the bat by 1.5m? (Don't forget to include kinetic energy, the bat is moving with the velocity you found in part C when it has been raised by 1.5m) - $W=\Delta KE+\delta PE=\frac{1}{2}\times2\times3.87^{2}+\frac{1}{2}\times0.5\times0.5\times0.1^{2}\times(\frac{3.87}{0.1})^2+\frac{1}{2}\times0.5\times4\times0.2^{2}\times(\frac{3.87}{0.2})^2+2\times9.8\times1.5=61.25\mathrm{\,​J}$ + $W=\Delta KE+\delta PE=\frac{1}{2}\times2\times3.87^{2}+\frac{1}{2}\times0.5\times0.5\times0.1^{2}\times(\frac{3.87}{0.1})^2+\frac{1}{2}\times0.5\times4\times0.2^{2}\times(\frac{3.87}{0.2})^2+2\times9.81\times1.5=61.25\mathrm{\,​J}$ ===== Question 3 Solution ===== ===== Question 3 Solution =====