A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 025 n/m and pushed downward. After the block is released, the spring will travel upward to the maximum height of 10.86 meters.
The relation between the potential energy on a string and the length it stretches is given by:
E = 1/2 . kx²
Where:
k = spring constant
x = spring stretch or compression
In the given problem:
k = 5,025 N/m
x = 0.106 m
Hence,
E = 1/2 . 5025 . (0.106)²
m . g . h = 1/2 . 5025 . (0.106)²
Where;
m = mass
h = height from the release point.
Hence,
0.260 x 10 x h = 28.23
h = 10.86 m
Therefore, the spring will travel upwards up to 10.86 m
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