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• Notes

$\begin{array}{l}{\text { 14 - 43. Determine the power input for a motor necessary }} \\ {\text { to lift 300 1b at a constant rate of 5 } \mathrm{ft} / \mathrm{s} \text { . The efficiency of the }} \\ {\text { motor is } \varepsilon=0.65 \text { . }}\end{array}$

$w=300l b$

$v=5 f t / s$

$\varepsilon=0.65$

$P=F.v=w \cdot v=30 0(5)=1500 ft.lb/s$

$\varepsilon=\frac{ \text {Power input }}{\text { power output }} \Rightarrow \text { power input }=\frac{\text { power output }}{\varepsilon}$

$=\frac{1500}{0.65}=2307.7 ft.lb/{s}$

$\begin{array}{l}{\text { 14- 78. The spring has a stiffness } k=200 \mathrm{N} / \mathrm{m} \text { and an }} \\ {\text { unstretched length of } 0.5 \mathrm{m} \text { . If is attached to the } 3-\mathrm{k} \text { g }} \\ {\text { smooth collar and the collar is released from rest at } 1 \text { . }} \\ {\text { determine the spect of the collar when it reaches } B \text { . Neglect }} \\ {\text { the size of the collar. }}\end{array}$

$K=200 \mathrm{N} / \mathrm{m}$

$l=0.5 \mathrm{m}$

$m=3 k g$

$\left(v_{g}\right)_{A}=\left(w h_{A}=m \cdot g{h_{A}}\right)=3(9.81)(2)=58.86 J$

$(v_ g)_{B}=m \cdot g{h_B}=m g 0=0$

$X_{A}=\sqrt{1.5^{2}+2^{2}}-0\cdot 5=2 m , \ X_{B}=1.5-0.5=1m$

$\left(V_{e}\right)_{A}=\frac{1}{2} k X_{A}^{2}=\frac{1}{2} 200(2)^{2}=400 J$

$\left(v_{e}\right)_B=\frac{1}{2} K X_{B}^{2}=\frac{1}{2} 200\left(1^{2}\right)=100 \mathrm{J}$

$T_{ A}+V_{A}=T_{B }+ V_{B}$

$0+(58.86+400)=\frac{1}{2} m V_B^{2}+(0+100 )$

$V_{B} \simeq 15.5 \mathrm{m} / \mathrm{s}$