Wagon in the figure can roll down the hill without friction. When the outpot speed (Vo) is 0 m/s and the end speed (v) is 4m/s
How high is the hill?
[WAGON] Vo = 0m/s
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_ [WAGON] Vo =4m/S
+14913 Wagon in the figure can roll down the hill without friction. When the outpot speed (Vo) is 0 m/s and the end speed (v) is 4m/s
How high is the hill?
\(E=\frac{m}{2}(v^2-v_o^2)=mgh\)
\(v^2-v_0^2=2gh \)
\((4\frac{m}{s})^2=2\cdot 9.81\frac{m}{s^2}\cdot h\)
\(h=16\cdot \frac{m^2}{s^2}/2\cdot9.81\cdot\frac{m}{s^2}\)
\(h=\frac{16\frac{m^2}{s^2}}{2\cdot 9.81\frac{m}{s^2}}\)
\(h=0.815\ m\)
\(The \ hill \ is \ 0.815 \ m \ high.\)
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+33615 Use conservation of energy:
potential energy at the top = m*g*h where m is mass, g is acceleration of gravity and h is height.
kinetic energy at the bottom = (1/2)mv2 where v is velocity
Equating the two we get
gh = (1/2)v2 or h = (1/2)v2/g
You can crunch the numbers!
Looks like asinus has already crunched the numbers for you!
