1.

A body of mass m=3.90 kg slides on a horizontal frictionless table with a speed of v=120 `ms^(-1)`. It is brought to rest in compressing a spring in its path. How much does spring is compressed, if its force constant k is 135 `Nm^(-1)`?A. 0.204 mB. 0.408 mC. 0.804 mD. 4.04 m

Answer» Correct Answer - A
(a) Given, m=3.9 kg, v=1.20 `ms^(-1)`
and k for spring=135 `Nm^(-1)`
From law of conservation of energy,
`(1)/(2)mv^(2)=(1)/(2)kx^(2)rArrx=sqrt((mv^(2))/(k))=sqrt((3.9xx1.2xx1.2)/(135))=0.204 m`


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