# Mechanical Energy:Potential, kinect, etc.

1. Dec 5, 2013

### Patolord

1. The problem statement, all variables and given/known data

A sphere of mass 2kg is release from point A off the edge of a hole that is represented of as a semi-circle of R=0,2 m.
consider g=10m/s²
What is the force that the ground apply on the sphere when it passes trough point B(bottom of the hole?

2. Relevant equations
Ep=m*g*h
Ec=m*v²/2
Q=m*v
F=m*a
3. The attempt at a solution
I tried finding the Potential energy of the sphere at point A that is
Ep=2kg*0,2m*10m/s² = 4J
and the weight of the sphere that is 2*10= 20 N
I don't know what to do anymore the velocity at point B i considered 2m/s because of
Ep=Ec=m*v²/2 then V=2m/s
Help?
Image not from actual problem point A would be the left point B where it is on the pic

Last edited: Dec 5, 2013
2. Dec 5, 2013

### ShayanJ

You can simply find the speed of the sphere when it reaches point B using conservation of energy.Call it $v_B$. When the sphere is moving in the semi-circle,it needs a centripetal force which at point B is equal to $\frac{mv_B^2}{R}$. But this centripetal force can be provided only by the circular track and nothing else so your answer is $\frac{mv_B^2}{R}$.

3. Dec 5, 2013

### Patolord

AH thanks the answer is 60N i guess you have to sum that with the weight that is 20N
40 +20 =60 N
but i dont get where mv2/r comes .. i dont have that on my theory
they just gave mv²/2 and m*g*h

4. Dec 5, 2013

### PhanthomJay

You need to understand the centripetal force concepts before solving this problem. You should know that the centripetal force mv^2/r is the net force acting inward toward the center of the circle. So the centripetal force is 40 N inward and is comprised of the net force of the ground normal force acting up
and the weight acting down, the net force

5. Dec 5, 2013

### Patolord

Thank you, now i see why couldnt soulve any problems envolving circles xD