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)`?

Options

A. 0.204 m
B. 0.408 m
C. 0.804 m
D. 4.04 m

Answer

A

Explanation

(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`