The potential energy stored in a spring stretched through a distance 'x' is given by:
U = (1/2)kx^2
where k is the spring constant.
In this problem, we are given that the potential energy of the spring when stretched through a distance 'a' is 25 J. Therefore, we can write:
25 = (1/2)ka^2
Solving for k, we get:
k = (2*25)/a^2 = 50/a^2
Now, we need to find the amount of work done on the spring to stretch it by an additional distance '5a'. The work done can be found using the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy plus the change in its potential energy.
In this case, there is no change in kinetic energy since the spring is not moving. Therefore, the work done is equal to the change in potential energy:
W = U_final - U_initial
where U_final is the potential energy of the spring when stretched through a distance '6a' (i.e., 'a' plus an additional '5a'), and U_initial is the potential energy of the spring when stretched through a distance 'a'.
The potential energy of the spring when stretched through a distance '6a' is:
U_final = (1/2)k(6a)^2 = (1/2)(50/a^2)(36a^2) = 900 J
Therefore, the amount of work done on the spring to stretch it by an additional distance '5a' is:
W = U_final - U_initial = 900 J - 25 J = 875 J
So, the amount of work done on the spring is 875 J.