A 0.290 kg block on a vertical spring with a spring constant of 5.00 ✕ 103 N/m is pushed downward, compressing the spring 0.110 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release? ___m

The height at point of release is 10.20 m

Explanation:

Given:

Spring constant : K= 5 x 10 to the 3rd power n/m

compression x = 0.10 m

Mass of block m= 0.250 kg

Here spring potential energy converted into potential energy,

mgh = 1/2 kx to the 2 power

For finding at what height it rise,

0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power

h= 10.20

Therefore, the height at point of release is 10.20 m

15.3 cos 65 degrees trig

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