Que 15. Two springs have their force constants as K1 and K2 (K1 > K2). On which spring is work done (i) when their lengths are increased by the same amount (ii) when they are stretched by the same force?
Ans- If each spring is stretched by x1 then,
work done on the first spring= ½ K1 X2
and work done on the second spring= ½ K2 x22 As K1 so, work done on the first spring is more.
Again, if same force say F be applied to the spring then, the extension in the first spring = x1=F/K1 and the extension is the second spring = x2= F/ K2 Thus work done on the
Second spring=F2/2K1 and work done on the second spring =F2/2K2
So, we find in this case work done is more for stretching the second spring.
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