1.

A spring is hanging from a rigid support. A force of 2000 dynes applied verticlly downwards at the lower end of the spring, extends it by 5 cm. The work done is given by:

Answer»

`5xx10^(3)` ERG
`5xx10^(4)` erg
`10^(6)` erg
1 Joule

Solution :`W=1/2kx^(3)=1/2(F/x)x^(2)`
=`1/2F.x.=1/2xx2000xx5=5000` ERGS`


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