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Quick question regarding kinetic energy, potential energy and velocity. 
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#1
Nov2212, 01:01 PM

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Ok so I have got the right answer to my calculation however I do not understand part of it,
here is the question...... Q2: The rollercoaster car from Task 1 Q2 above is initially travelling at 6.5m/s at point A in the diagram below. Assuming negligible frictional forces as previously, what will be the final velocity of the car at point B? The image shows that the cart drops by 3.5 m on a ramp, and point b is further along the track. So here is what I did: At the top of the drop the drop the carriage already has an amount of kinetic energy, and potential energy, the potential energy will add on to the kinetic energy as it begins to accelerate down the drop. Initial Kinetic Energy = 0.5 x mass x velocity^2 = 0.5x440x6.5^2 = 9295 joules Potential energy = mass x gravity x height = 440x9.8x3.5 = 15092 joules Total Energy at bottom of drop = 15092+9295= 24387 joules. Transpose this formula to find V: KE= 0.5 x mass x v^2 V = √2xKinetic Energy/Mass V = √2x24387/440 V = 10.53 m/s Now here is my question why is it that I have to calculate the energies to find the velocity rather than calculate the velocities individually, i.e. find the velocity of the carriage dropping using the initial velocity as 0. 


#2
Nov2212, 01:29 PM

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Because energy depends on the square of the velocity ... what you are doing is working out the kinetic energy gained in the drop, and, from that, deducing the final speed.
Work it out symbolically instead of numerically and you'll see. 


#3
Nov2212, 02:07 PM

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Thanks for that, however any chance you could slightly expand on that?



#4
Nov2212, 02:10 PM

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Quick question regarding kinetic energy, potential energy and velocity.
If the slope of the ramp is a straight line, you can use Newton's laws and the equations for kinematics at constant acceleration to find the final velocity, and you should get the same result. Try it, it's a good exercise!
If the slope isn't a straight line, then you can't do this (at least not easily!) because the acceleration isn't constant. But conservation of energy still works. 


#5
Nov2212, 02:22 PM

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Yeah the diagram doesn't state whether its straight or not, cheers though.



#6
Nov2212, 02:28 PM

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You should (as well as working the problem for a constant slope ramp) try to express the idea you are asking questions about symbolically ... you know, with letters instead of numbers. Compare with the same thing using the proper conservation of energy approach. eg. for conservation of energy you find: ##mgh+\frac{1}{2}mv_i^2=\frac{1}{2}mv_f^2## which gives you: ##v_f=\sqrt{2gh+v_i^2}## Now you can see the relationship more clearly. See if you can express the alternative idea you talked about as a relation and solve it for ##v_f## using only the symbols. [edit] 


#7
Nov2212, 03:17 PM

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Cheers for that, think I can get my head round it now.



#8
Nov2212, 05:10 PM

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