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. 0.0375 J, 0.125 J
B. 0.125 J, 0.375 J
C. 0.125 J, 0.125 J
D. 0.375 J, 0.375 J

Answer

A

Explanation

Here, `m=1kg,k=100" N "m^(-1)`
A=10cm=0.1m
The blocks executes SHM, its angular frequency is given by
`omega=sqrt((k)/(m))=sqrt((100" N "m^(-1))/(1kg))=10" rad "s^(-1)`
velocity of the block at x=5cm=0.05m is
`v=omegasqrt(A^(2)-x^(2))=10sqrt((0.1)^(2)-(0.05)^(2))=10sqrt(7.5xx10^(-3))ms^(-1)`
Kinetic energy of the block,
`K=(1)/(2)mv^(2)=(1)/(2)xx1xx0.75=0.0375J`
Potential energy of the block,
`U=(1)/(2)kx^(2)=(1)/(x)xx100xx(0.05)^(2)=0.125J`