A mass of 0.3 kg hangs motionless from a vertical spring whose length is 0.77 m and whose unstretched length is 0.59 m. Next the mass is pulled down to where the spring has a length of 0.84 m and given an initial speed upwards of 1.8 m/s. What is the maximum length of the spring during the motion that follows?

0.93m

Explanation:

Getting the data:

1. mass (M): 0.3 kg

2. Length with the mass (L1): 0.77 m

3. Unstretched length (Lu): 0.59 m

4. Length after the pulling (Lf): 0.84 m

5. Speed upwards (): 1.8 m/s

By laws of newton and with the first and second data we will find the spring constant using:

M*g = kx

Where g is the gravity, M*g is the weight of the mass, k is the spring constant and x is the stretch (Lu-L1), so:

N/m

Now using the conservation of energy:

Where E1 is the energy when the length of the spring is 0.84m, = Lf-Lu,  E2 is the energy when the length is maximun and is the stretch when the length is maximun. So:

it means that the maximun length of the spring is:

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

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