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# Physics (energy)

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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

Guest May 9, 2017
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#1
+7059
+1

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?

$$E=\frac{m}{2}(v^2-v_o^2)=mgh$$

$$v^2-v_0^2=2gh$$

$$(4\frac{m}{s})^2=2\cdot 9.81\frac{m}{s^2}\cdot h$$

$$h=16\cdot \frac{m^2}{s^2}/2\cdot9.81\cdot\frac{m}{s^2}$$

$$h=\frac{16\frac{m^2}{s^2}}{2\cdot 9.81\frac{m}{s^2}}$$

$$h=0.815\ m$$

$$The \ hill \ is \ 0.815 \ m \ high.$$

!

asinus  May 9, 2017
edited by asinus  May 9, 2017
#2
+26236
+1

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!

Alan  May 9, 2017
edited by Alan  May 9, 2017

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