A block of mass 5.0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquired an upward speed of 2.0 m/s . How high will it rise?

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Solution

Given in the question :-
Mass of the block = 5 kg.
v = 2m/sec.
Now Kinetic energy of the block is ,
(1/2) mv² .
Assume the new height of the block be h.
So the change in potential energy at the highest point is equal to Kinetic energy .
∴ mgh = (1/2)mv²
h = (1/2g) v²
h = 2² /(2× 10)
h = 1/5
h = 0.2 m
or h = 20 cm.

It will rise upto 20 cm.

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