A block of mass 4 kg hangs from a spring constant `k=400(N)/(m)`. The block is pulled down through 15 cm below and released. What is its kinetic energy when the block is 10 above the equilibrium position.?

Options

A. `5J`
B. `2.5J`
C. `1J`
D. `1.9J`

Answer

B

Explanation

Amplitude `=0.15m`
`omega=sqrt((k)/(m))=sqrt((400)/(4))=10(rad)/(s)impliesf=(omega)/(2pi)=(5)/(pi)Hz`
Energy of the particle executing `SHM=KE+PE`
`=(1)/(2)mv^2+(1)/(2)kx^2=(1)/(2)kA^2`
Therefore, `(1)/(2)mv^2=(1)/(2)kA^2-(1)/(2)kx^2=(1)/(2)k(A^2-x^2)`
`=(1)/(2)xx400(0.15^2-0.1^2)=2.5J`