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• Notes

$\begin{array}{l}{\text { How many joules of energy does a } 100 \text { -watt light bulb use }} \\ {\text { per hour? How fast would a } 70-\text { kg person have to run to have that }} \\ {\text { amount of kinetic energy? }}\end{array}$

$P=\frac{W}{\Delta t}$

$\rightarrow W=P \Delta t=100 * 3600=3.6 * 10^{5} \mathrm{J}$

$k=w=3.6 * 10^{5} J$

${K=\frac{1}{2} \mathrm{mv}^{2} \longrightarrow \mathrm{V}=\sqrt{\frac{2 \mathrm{K}}{\mathrm{m}}}=\sqrt{\frac{2\left(3.6 \times 10^{5}\right)}{70}}=100 \mathrm{m} / \mathrm{s}}\$

$\begin{array}{l}{\text { An elevator has mass } 600 \mathrm{kg} \text { , not including passengers. }} \\ {\text { The elevator is designed to ascend, at constant speed, a vertical }} \\ {\text { distance of } 20.0 \mathrm{m} \text { (five floors) in } 16.0 \mathrm{s} \text { , and it is driven by a }} \\ {\text { motor that can provide up to } 40 \mathrm{hp} \text { to the elevator. What is the }} \\ {\text { maximum number of passengers that can ride in the elevator? }} \\ {\text { Assume that an average passenger has mass } 65.0 \mathrm{kg} \text { . }}\end{array}$

$\uparrow\text {up }, \quad t=16 \text { s, comstant speed, } 65 \mathrm{kg}$

$P_{a vg}=\frac{\Delta w}{\Delta t}=\frac{mgh}{t} \longrightarrow m=\frac{P_{av}t}{gh}$

$P_{av}=\text { 40hp }\left(\frac{746 \mathrm{w}}{1 \mathrm{hp}}\right)=2.984*10^{4} \mathrm{w}$

$m=\frac{2.984 * 10^{4} * 16}{9.8 * 20 m}=2.436+10^{3} kg$

$m_{\text {all pass. }}=2.436 * 10^{3}-600=1.836 * 10^{3} \mathrm{kg}$

$\text {# passengers}= \frac{1.836*10^{3}}{65}=28.2$