Kinetic Energy of a mass moving in horizontal circle

  • #1

Homework Statement


A mass moves in a circular path that has a radius of 24.6cm on a horizontal frictionaless surface. If the centripetal force acting on the mass is 96.5N, what is the kinetic energy of the mass?
r=0.246m
Fc=96.5N


Homework Equations


He told us to use these and "play around with them":
Fc=mv2/r
Ek=0.5mv2



The Attempt at a Solution


I have not a clue as to how to even begin this one. Maybe use the idea that acceleration centripetal=force centripetal(minus mass)? So use:
ac=v2/r
96.5N=v2/0.246m
v2=392m/s
v=19.8m/s

Fc=mv2/r
96.5N(0.246m)=m(19.8m/s)2
m=0.0605kg

Ek=0.5mv2
Ek=0.5(0,0605kg)(392m^2/s^2)
Ek=11.9J

Or am I completely and irrevocably lost?
 

Answers and Replies

  • #2
92
0
This has been staring me in the face for a while...

I now think I have the solution. Look at your two equations, and look what they have in common. Once you find one, you can use it to find the Kinetic Energy of that mass. Good Luck!
 
  • #3
Just to update, I got that question correct. In the solution the teacher showed us, he combined the two formulas and got the same answer.
 

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