A 153 g mass is attached to the end of an unstressed vertical spring (of constant 24.7 N/m) and then dropped. The acceleration of gravity is 9.8 m/s 2 . What is its maximum speed

The answer to the question is

Its maximum speed is 1.54 m/s

Explanation:

Work done = Kinetic energy

0.5·m·v² = 0.5·k·x²

Where

m = mass

v = velocity

k =  spring constant

x = extension of the spring

We note that Force F is given by

F = m·a

Where

a = acceleration due to gravity

= 0.153×9.8 = 1.4994 N

Equating the work done by the force to the work done on the spring gives

Work done = Force × Distance = 1.4994×x = 0.5×k÷x² = 0.5×24.7×x²

x = 1.4994÷12.35 = 0.121 m

Substituting the value of x into the equation below gives

0.5·m·v² = 0.5·k·x²

0.5×0.153×v² = 12.35×0.121²

v² = 0.182÷0.0765 = 2.379

v = 1.54 m/s

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vapour or gas assuming atmospheriv pressure