If the potential energy of a spring when stretched through a distance ‘a’ is 25 J, then what is the amount of work done on the same spring so as to stretch it by an additional distance ‘5a’?
P. E = 1 2 2 ka 1 2 2 kx = 25 Additional distance of 5a becomes 6a Substituting ‘a’ we get = 900 J Additional work done = 900 – 25 = 875 J
