1.

A spring is stretched by 3 cm, its potential energy is U_1. If the spring is stretched by 6 cm, its potential energy be that is stored in it is U_(2)="……"U_(1). (Fill in the blank).

Answer»

Solution :`U_(1) = (1)/(2)ky_(1)^(2)= (1)/(2)k(3)^(2)= (1)/(2)KXX 9`
`U_(2)= (1)/(2)ky_(2)^(2)= (1)/(2)k(6)^(2)= (1)/(2)kxx 36`
`therefore (U_2)/(U_1)= (36)/(9)= 4`
`therefore U_(2) = 4U_(1)`.


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