Question

n to the Owens Valley 1632 m below. If the 21.0 m along the road to the top of the hill. If the cyclist's initial speed is 9.30 m/s, what is the final speed? Ignore friction and air

Answer 1

Initial energy at the top = E_i = mgh final energy at the bottom E_f = 1/2 mv^2 total energy is conserved E_f = E_i v = squareroot 2 gh = squareroot 2 times 9.8 times 1632 = 178.85 m/s Initial energy at the bottom E_i = 1/2 mv_i^2 final energy at the top E_f = KE_f + PE_f = 1/2 mv_f^2 + mgh h = L sin theta E_f = 1/2 mv_f^2 + mg L sin theta total energy is conserved E_f = E_i v_f = squareroot v_i^2 - 2 gL sin theta = squareroot 9.3^2 - 2 times 9.8 times 21 sin 11.1 = 2.69 m/s Initial nergy at the top E_i = mgh final energy at the bottom E_f = 1/2 KX^2 total energy is conserved E_f = E_i x = squareroot 2 mgh/k = squareroot 2 times 4.3 times 9.8 times 2.5/2.38 times 10^4 = 0.094 m =

$\\8) \\Initial\;energy\;at\;the\;bottom\;E_i=\frac{1}{2}mv_i^2\\\\\\ final\;energy\;at\;the\;top\\\\\\E_f=KE_f+PE_f=\frac{1}{2}mv_f^2+mgh\\\\ h=L\;sin\theta\\\\ E_f=\frac{1}{2}mv_f^2+mgLsin\theta\\\\\\ total\;energy\;is\;conserved\\\\\\E_f=E_i\\\\\\ v_f=\sqrt{v_i^2-2gLsin\theta}=\sqrt{9.3^2-2\times 9.8\times 21\sin11.1}=2.69\;m/s$ $\\Initial\;energy\;at\;the\;top\;E_i=mgh\\\\\\ final\;energy\;at\;the\;bottom\\\\\\E_f=\frac{1}{2}kx^2\\\\\\ total\;energy\;is\;conserved\\\\\\E_f=E_i\\\\\\ x=\sqrt{\frac{2mgh}{k}}=\sqrt{\frac{2\times 4.3\times 9.8\times 2.5}{2.38\times 10^4}}=0.094\;m=9.4\;cm$

$\\9)\\\\(a)\\ Potential \;energy\;stored\;in\;bow=kinetic\;energy\;of\;the\;arrow\\\\\\ PE=\frac{1}{2}mv^2=\frac{1}{2}5.25\times 10^{-2}\times 39^2=40\;J\\\\\\ (b)potential\;energy\;PE=\frac{1}{2}kx^2\\\\\\averge\;force\;F_{avg}=\frac{1}{2}kx=\frac{PE}{x}\\\\F_{avg}=\frac{40}{0.3}=133.3N$