1.

The pendulum shown in the figure is kept horizontal. It is released from this positions. What is its velocity when it reaches the lowest position? (g=9.8ms^(-2))

Answer»


Solution :Kinetic energy at POSITION B
= potential energy at position A
`(1)/(2)mv_(max)^(2)=mgh`
`therefore v_(max)^(2)=2gh`
`therefore v_(max)=sqrt(2gh)`
`therefore` Maximum velocity of PENDULUM,
`v_(max)=sqrt(2xx9.8xx2.5)`
`=sqrt(49)`
`=7ms^(-1)`


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