Respuesta :
Answer: elastic potential energy = 20.27 J
Explanation:
Given that the
Mass M = 0.470 kg
Height h = 4.40 m
Spring constant K = 85 N/m
The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.
But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.
That is
K .E = P.E = mgh
Where g = 9.8m/s^2
Substitutes all the parameters into the formula
K.E = 0.470 × 9.8 × 4.4
K.E = 20.27 J
Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.
The maximum potential energy stored in the spring during the motion of the blocks is 20.39 J.
The given parameters;
- mass of the block, m = 0.47 kg
- spring constant, k = 85 N/m
The extension of the spring is calculated as follows;
[tex]F =kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{0.47 \times 9.8}{85} \\\\x = 0.054 \ m[/tex]
The maximum potential energy stored in the spring during the motion of the blocks is calculated as;
[tex]P.E = \frac{1}{2} kx^2 \ + mgh\\\\P.E = \frac{1}{2} (85)(0.054)^2 \ + \ (0.47)(9.8)(4.4)\\\\P.E = 20.39 \ J[/tex]
Thus, the maximum potential energy stored in the spring during the motion of the blocks is 20.39 J.
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