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)(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Kinetic and Potential Energy calculation

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