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Kinetic and Potential Energy calculation

  • Thread starter Blubla
  • Start date
5
0
A 215 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 30.0 cm. Point B to C have Height= 2/3(R)
Calculate the following: A. The gravitational potential energy at A relative to B
B. The particle's kinetic energy at B
c. Th particle's speed at B
D. The potential energy and kinetic energy at C

I solved A,B,C and D(The potential Energy but not the kinetic energy)

Here is my work:

A. Changed 215 g into Kg so it will be 0.215 kg
30.0 cm into m so it will be 0.30m
PE=mgh
PE=0.215*9.81*0.30
PE=0.633J



B. KE= 1/2mv^2
KE=1/2*0.215*2.4^2
KE=0.62J


C. V^2f=2gh
V^2f=2*9.81*030
V^2f=5.9 Apply square root
Vf= 2.4m/s

D. PE=mgh
PE=0.30*9.8(2/3*0.30)
PE=0.422

Can someone please help me find the KE at C, the answer is 0.211J according to the book. Well, what I got is 0.43J..



Thanks in advance.
 

Answers and Replies

You missed the point of the problem. Use conservation of energy.
 
That is true, you need to use the conservation of energy between C and B.
mgR= KEc + mg(2/3)R
KEc=1/3mgR=0.211J
 

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