A spring is streach by 0.20m when mass of 0.50kg is suspended when a mass of 0.25kg is suspended

Answer 1

According to Hooke's Law, the force applied to a spring is directly proportional to its extension. Therefore, if the mass is halved, then the force would be halved, so the extension would be halved.

Answer 2

The data on a spring is clear, but there is no question. So, I will assume that what's wanted is how is the spring extended when a mass of 0.25 kg is attached. The equation to use if F = kx, where F is the force, k is the spring constant and x is the distance stretched. Now F is NOT 0.5 kg in the first instance. F = mg with g = about 10 m/s^2. Plugging in and solving we get k = 25 N/m. Using the same equation and solving for x we get F/k and plugging in for the second situation we get x = 0.1 m.

Of course this is long story of it. The short story is that since the F is halved by reducing the weight by a factor of two, so is x. This is a consequence of the F = kx being linear, meaning that changes in x are directly proportional to changes in F.