A block whose mass is 2 kg is fastened on a spring whose spring constant is 100 `Nm^(-1)` . It is pulled to a distance of 0.1 m from over a frictionless surface and is released at t=0. Calculate the kinetic eneryg of the block when it is 0.05 m away from its mean position.

Explanation

The block executes SHM,its angular frequency.
`omega =sqrt((k)/(m)) = sqrt((100Nm^(-1))/(2kg))= 7.07 rad s^(-1)`
Its displacement at any time t is
`x(t)= acos omegat`
`= 0.1 cos ( 7.07t)`
When the particle is 0.05 m away from the mean position , `0.05 = 0.1 cos ( 7.07t )`
or `cos( 7.07t) = 0.5`
or `sin(7.07 t) = (sqrt(3))?9 2) = 0.866`
Velocity of the block at `x= 0.05 m` is
`v= A omega sin omegat`
`= 0.1 xx 7.07 xx 0.866`
`= 0.61 ms^(-1)`
Hence, `K.E.= (1)/(2) mv^(2)`
`= (1)/(2) xx 2(0.61)^(2)`
`=0.37 J`