A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 100N/m. the block is pulled to a distance x=10 cm from its equilibrium position at x=0 on a frictionless surface from rest at t=0. the kinetic energy and potential energy of the block when it is 5 cm away from the mean position is

Options

A. 0.375 J, 0.125 J
B. 0.125 J, 0.375 J
C. 0.125 J, 0.125 J
D. 0.375J, 0.375J

Answer

A

Explanation

Given, `m=1kg,k=100N//m`
`A=10cm=0.1m`
Angular frequency of SHM of the block is
`omega=sqrt((k)/(mm))=sqrt((100Nm^(-1))/(1kg))=10` rad/s
`therefore`velocity of the blocka t `x=5cm`
`=0.05m` is
`v=omegasqrt(A^(2)-x^(2))`
`=10sqrt((0.1)^(2)-(0.05)^(2))`
`=10sqrt(7.5xx10^-3)m//s`
Kinetic energy of the block.
`K=(1)/(2)mv^(2)=(1)/(2)xx1xx0.75`
`=0.375J`
Potential energy of the block.
`U=(1)/(2)kx^(2)=(1)/(2)xx100xx(0.05)^(2)`
`=0.125J`