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Find maximum kinect energy 
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#1
Jan113, 08:42 AM

P: 33

Hi,
I am trying to solve a problem but I can't figure how to continue.. There's an object moving in a circular path with radius 0.5m in an horizontal surface. The rope will break if the tension exceeds 16N. What is the maximum kinect energy? So r = 0.5m T = 16N K=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex] How can I find m and v? I tried with T = mg <=> 16 = m(9.8) <=> m = 1.63kg but I don't know if it is correct. But besides that I can't figure how to find v 


#3
Jan113, 08:54 AM

P: 33

I also tried that
v = [itex]\sqrt{a*r}[/itex] But how to find a? Will it be g? 


#4
Jan113, 08:55 AM

P: 154

Find maximum kinect energy
yes..



#5
Jan113, 09:01 AM

P: 33

It is solved :) Thanks!



#6
Jan113, 09:03 AM

Mentor
P: 41,436




#7
Jan113, 09:12 AM

P: 33




#8
Jan113, 09:22 AM

Mentor
P: 41,436

Hint: You don't need the mass to answer the question. It asks for kinetic energy, not mass. 


#9
Jan113, 09:41 AM

P: 33

<=> 16 = 1.63 * [itex]\frac{v^{2}}{0.5}[/itex] But since 1.63 isn't right it will give wrong results But K=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex] so I need the mass 


#10
Jan113, 09:46 AM

Mentor
P: 41,436




#11
Jan113, 10:01 AM

P: 33

So, i did this: F = ma <=> a = [itex]\frac{F}{m}[/itex] a = [itex]\frac{v^{2}}{r}[/itex] <=> [itex]v^{2}=a*r[/itex] <=> [itex]v^{2}=\frac{F}{m}*r[/itex] K = [itex]\frac{1}{2}*m*\frac{F*r}{m} = \frac{1}{2}*F*r[/itex] Thanks! 


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