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• Notes

$\begin{array}{l}{\text { Adult cheetahs, the fastest of the }} \\ {\text { great cats, have a mass of about } 70 \mathrm{kg} \text { and have been clocked run- }} \\ {\text { ning at up to } 72 \mathrm{mph}(32 \mathrm{m} / \mathrm{s}) . \text { (a) How many joules of kinetic }} \\ {\text { energy does such a swift cheetah have? (b) By what factor would }} \\ {\text { its kinetic energy change if its speed were doubled? }}\end{array}$

$m=70kg \quad, \quad V=32 \mathrm{m/s}$

$k=\frac{1}{2} m v^{2}$

(a) $k=\frac{1}{2}(70)(32)^{2}=3.6 \times 10^{4} J$

$K=\frac{1}{2}(70)(64)^{2}=14.4 \times 10^{4} J$

$\frac{3.6 * 10^{4}}{14.4 *10^{4}}=\frac{1}{4} \longrightarrow 4 \text { times }$

$\begin{array}{l}{\text { Use the work-energy theorem to solve each of these prob- }} \\ {\text { lems. You can use Newton's laws to check your answers. Neglect }} \\ {\text { air resistance in all cases. (a) A branch falls from the top of a }} \\ {95.0 \text { -m-tall redwood tree, starting from rest. How fast is it moving }} \\ {\text { when it reaches the ground? (b) A volcano ejects a boulder directly }}\end{array}$

$\begin{array}{l}{\text { upward } 525 \mathrm{m} \text { into the air. How fast was the boulder moving just }} \\ {\text { as it left the volcano? (c) A skier moving at 5.00 } \mathrm{m} / \mathrm{s} \text { encounters a }} \\ {\text { long, rough horizontal patch of snow having coefficient of kinetic }} \\ {\text { friction } 0.220 \text { with her skis. How far does she travel on this patch }} \\ {\text { before stopping? (d) Suppose the rough patch in part ( } \mathrm{c}) \text { was only }}\end{array}$

$\begin{array}{c}{2.90 \mathrm{m} \text { long? How fast would the skier be moving when she }} \\ {\text { reached the end of the patch?(e) At the base of a frictionless icy }} \\ {\text { hill that rises at } 25.0^{\circ} \text { above the horizontal, a toboggan has a speed }} \\ {\text { of } 12.0 \mathrm{m} / \mathrm{s} \text { toward the hill. How high vertically above the base }} \\ {\text { will it go before stopping? }}\end{array}$

(a) $k_{1}=0, V_{1} = 0$

$W_{\text {total }}=W_{g r av}=m g s$

$K=w \Rightarrow {mg} s=\frac{1}{2} m{v_2}^{2}$

$\longrightarrow V_{2}=\sqrt{2 g s}=\sqrt{2(9.8)(95)}=43.2 \mathrm{m/s}$

(b) $V_{2}=0 \rightarrow K_{2}=0$

$w_{\text {total }}=w_{\text {grav }}=-m g s$

$k_{1}=-m g s \rightarrow \frac{1}{2} m {v_{1}}^{2}=-m g s \rightarrow v_{1}=\sqrt{2gs}$

$\longrightarrow V_{1}=\sqrt{2(9.8)(525)}=101 \mathrm{m/s}$

(c) $k_{2}=0$

$W_{\text {total }}=W_F=-\mu_k mg s$

${k_{1}=\frac{1}{2} m {v_{1}}^{2}=-\mu_ k \text { mgs } \rightarrow s=\frac{V_{1}^{2}}{2 \mu_{k} g}=\frac{(5)^{2}}{2 * 0.22 * 98}}\$

$\rightarrow S=5.8 \mathrm{m}$

(d) $K_{1}=\frac{1}{2} m {v_{1}}^{2}, K_{2}=\frac{1}{2} m{ v_{2}}^{2}, W_{tot}=W_{f}=-\mu_{k} m g{s}$

$w_{\text {total }}=w_{f}=\mu_k mgs$

$k_{2}=w_{tot}+k_{1} \Rightarrow \quad \frac{1}{2} m {v_2}^2=-\mu_{k}m g s+\frac{1}{2} m{v_1}^{2}$

$V_{2}=\sqrt{V_{1}^{2}-2\mu_{k} g s} \longrightarrow \sqrt{5^{2}-2(0.22)(9.8)(2.9)}=3.53 \mathrm{m/}s$

(e) $k_{2}=0 \rightarrow v_{2}=0 \quad, \quad K_{1}=\frac{1}{2} m {v_{1}}^{2}, \quad w_{grav}=-m g y?$

$k_{1}=-m g y_{2} \rightarrow \frac{1}{2} m{ v_{1}}^{2}=-mg y_{2}$

$\rightarrow y_{2}=\frac{{v_{1}}^{2}}{2 g}=\frac{(12)^{2}}{2(9.8)}=7.35 \mathrm{m}$

$\begin{array}{l}{\text { A soccer ball with mass } 0.420 \mathrm{kg} \text { is initially moving with }} \\ {\text { speed } 2.00 \mathrm{m} / \mathrm{s} \text { . A socer player kicks the ball, exerting a constant }} \\ {\text { force of magnitude } 40.0 \mathrm{N} \text { in the same direction as the ball's }} \\ {\text { motion. Over what distance must the player's foot be in contact }} \\ {\text { with the ball to increase the ball's speed to } 6.00 \mathrm{m} / \mathrm{s} \text { ? }}\end{array}$

$m=0.42 \mathrm{kg}, \quad v_{1}=2 \mathrm{m} / \mathrm{s} \quad, \quad F=40 \mathrm{N} \quad, \mathrm{v}_{2}=6 \mathrm{m} / \mathrm{s} \quad \mathrm{s} ? ?$

$W_{total}=k_{2}-k_{1}$

$k_{1}=\frac{1}{2} m {v_{1}}^{2}=\frac{1}{2}(0.42)(2)^{2}=0.84 J$

$k_{2}=\frac{1}{2} m {v_{2}}^{2}=\frac{1}{2}(0.42)(6)^{2}=7.56 \mathrm{J}$

$W_{t o tal}=k_{2}-k_{1}=7.56-0.84=6.72 J$

$W_{F}=FS \cos \theta$

$S=\frac{W_ F}{F \cos \theta}=\frac{6.72}{40* \cos 0}=0.168 \mathrm{m}$