A block of mass 4 kg is dropped from a height of 0.8 m on to a spring of spring constant k = 1960 N m⁻¹, Find the maximum distance through which the spring will be compressed. (Ans: 0.18 m)



Data Given:


Mass of the block = m = 4 kg

Height = h = 0.8 m

Spring constant = K = 1960 N m⁻¹

Value of acceleration due to gravity = g = 9.8 m s⁻²


To Find:


Maximum distance  = x = ?


Solution:


According to the law of conservation of energy

Loss of P.E = Elastic P.E. Store in spring

or 

mgh = `\frac {1}{2}` Kx²

or

x = `\sqrt frac {2mgh}{k}`

By putting values

x = `\sqrt frac {2ｘ4 kgｘ 9.8 m s⁻²ｘ0.8 m}{1960 N m⁻¹}`

x = `\sqrt {0.032 m²}`

x = 0.179 m

x = 0.18 m -----------------Ans.

Hence the spring will compress at a maximum distance of 0.18 m 



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