Answer :
The potential energy of the spring-mass system when the spring is only stretched by our distance of a/2 is 2.5 Joules.
The potential energy of a spring for a particular mass spring system is given by,
U = 1/2kx²
x is the extension in the spring,
K is the spring constant.
If the spring is stretched by a distance of a then the potential energy of the mass spring system,
10 = 1/2ka²
If the spring mass-system is only stretched by a distance of a/2, then let us say that potential energy of the system is U',
U' = 1/2k(a/2)²
U' = 1/2ka²/4
Putting, 1/2ka² = 10,
U' = 10/4
U' = 2.5 Joules.
So, the potential energy of the system is now 2.5 joules.
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