# If the potential energy of a spring when stretched through a distance ‘a’ is 25 J, then what is theamount of work done on the same spring so as to stretch it by an additional distance ‘5a’?

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Let the stiffness constant of the spring be k

Therefore, the elastic potential energy stored by the spring when it is stretched through a distance of 'a' from its natural length = (1/2) X k X a X a = 25 J (given).

So, k X a X a = 50 J

Now, the total elastic potential energy stored by the spring when it is stretched through an additional distance of '5a' that is the total distance of '6a' from its natural length = (1/2) X k X 6a X 6a

= 18 X k X a X a

= 18 X 50 J

= 900 J