Answer to Question #217307 in Classical Mechanics for Help

Question

Use the information below to answer the following 2 questions.

A rubber-band slingshot is drawn back is drawn back 0.15 m with a force of 28 N and shoots a 28-g stone. Solve using energy considerations. Recall from the previous lesson that E_p = ½ kx². Since F = kx, E_p = ½ Fx.

1)The initial speed of the stone if the rubber band is released is___m/s

2) The maximum height of the stone if it is shot straight up is ___m

Use the information below to answer the following 2 questions.

A 0.80-kg coconut is growing 10 m above the ground in its palm tree. The tree is just at the edge of a cliff that is 15 m tall.

1) The maximum speed of the coconut if it fell to the ground beneath the tree is ___m/s.

2) The maximum speed of the coconut if it fell from the tree to the bottom of the cliff is ___m/s

Answer 1

$1.$

$x = 0.15m$

$F = 28N$

$m= 28g= 0.028kg$

$g= 9.8\frac{m}{s^2}$

$E= E_{r}+E{s}$

$E_{r},E{s}-\text{energy of rubber-band slingshot and stone}$

$\text{before the shot:}$

$E=E_{pr}+E_{kr}+E_{ps}+E_{ks}$

$\text{energy of rubber-band slingshot}$

$E_{kr }=0$

$E_{pr }=\frac{1}{2}Fx=\frac{1}{2}*28*0.15= 2.1 J$

$\text{energy of stone}$

$E_{ks}=0;E_{ps}=0$

$E= 2.1J$

$\text{shot fired: }$

$E=E_{pr}+E_{kr}+E_{ps}+E_{ks} = 2.1J$

$E_{pr}=0;E_{kr}=0;$

$E_{ps}=0;$

$E_{ks}= \frac{mv^2}{2}$

$\frac{mv^2}{2}=E$

$v= \sqrt{\frac{2E}{m}}=\sqrt{\frac{2*2.1}{0.028}}=12.25\frac{m}{s}$

$\text{the stone has reached its highest point:}$

$E=E_{pr}+E_{kr}+E_{ps}+E_{ks}$

$E_{pr}=0;E_{kr}=0$

$E_{ps}= mgh;E_{ks}=0;$

$E = mgh$

$h = \frac{E}{mg}= \frac{2.1}{0.028*9.8}=7.65m$

$\text{Answer:}$

$\text{The initial speed of the stone if the rubber band is released is 12.25 m/s}$

$\text{The maximum height of the stone if it is shot straight up is 7.65 m}$

2.

$\text{Coconut falls from the tree:}$

$h = 10m$

$m = 0.8 kg$

$E= E_p+E_k$

$E_{pmax}= mgh=0.8*9.8*10= 78.4J$

$E_{pmax}= E_{kmax}$

$E_{kmax}=\frac{mv^2}{2}$

$v=\sqrt{\frac{2*E_{kmax}}{m}}=\sqrt{{\frac{2*78.4}{0.8}}}=14\frac{m}{s}$

$\text {Coconut falls from tree and cliff:}$

$h_1 = h+15= 25m$

$E'= E'_p+E'_k$

$E'_{pmax}= mgh_1=0.8*9.8*25= 196J$

$E'_{pmax}= E'_{kmax}$

$E'_{kmax}=\frac{mv_1^2}{2}$

$v_1=\sqrt{\frac{2*E'_{kmax}}{m}}=\sqrt{{\frac{2*196}{0.8}}}=22.16\frac{m}{s}$

$\text{Answer:}$

$\text{The maximum speed of the coconut if it fell to the ground }$

$\text{beneath the tree is 14m/s.}$

$\text{The maximum speed of the coconut if it fell from the tree}$

$\text{to the bottom of the cliff is 22.16m/s}$