+0  
 
+1
218
2
avatar

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

                           _

                                    _      

                                                  _ 

                                                                    _           [WAGON] Vo =4m/S

Guest May 9, 2017
Sort: 

2+0 Answers

 #1
avatar+7155 
+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?

 

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

 

\(v^2-v_0^2=2gh \)

 

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

 

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

 

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

 

\(h=0.815\ m\)

 

\(The \ hill \ is \ 0.815 \ m \ high.\)

 

laugh  !

asinus  May 9, 2017
edited by asinus  May 9, 2017
 #2
avatar+26357 
+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

7 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details