A block is released from position A in the following figure. Find the speed of the block in position B. Assume the surface as frictionless.

Explanation

(a) `K_A=0`,`U_A=mgR`
`K_B=(1)/(2)mv^2`,`U_B=mgRcostheta`
`K_A+U_A=K_B+U_B`
`v=sqrt(2gR(1-costheta))`
(b)
`K_A=0`,`U_A=mgRcosalpha`
`K_B=(1)/(2)mv^2`,`U_B=mgRcos(alpha+beta)`
`K_A+U_A=K_B+U_B`
`v=sqrt(2gR[cosalpha-cos(alpha+beta)])`
(c )
`K_A=0`,`U_A=mgxx5R=5mgR`
`K_B=(1)/(2)mv^2`,`U_B=mgR(1+cos60^@)=(3)/(2)mgR`
`K_A+U_A=K_B+U_B`
`0+5mgR=(1)/(2)mv^2+(3)/(2)mgR`
`v=sqrt(7gR)`