A block whose mass is 1kg is fastened to a spring. The spring has a spring constant of `50 Nm^(-1)`. 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. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

Explanation

`m = 1kg, k = 50 Nm^(-1), A= 10 cm = 0.1 m, x=5 cm = 0.05m`
Angular frequency of the block executes SHM, `omega = sqrt((k)/(m)) = sqrt((50)/(1)) =7.07 rad s^(-1)`.
Its displacement at any time t is, `x(t) = A cos omega t = 0.1 cos omega t`.
Now if `x= 5 cm =0.05 m` then `0.05 = 0.1 cos omega t`
`therefore 0.5 = cos (omega t)`
`therefore cos (omega t)= (1)/(2)`
`therefore sin omega t= sin (pi)/(3) = (sqrt(3))/(2)`
`sin omega t = 0.866 " ""........."(1)`
The velocity of the block at x= 5 cm is, `x(t) = A cos omega t`
`therefore v= (dx(t))/(dt) = (d(A cos omega t))/(dt)`
`therefore v= -A omega sin omega t`
`= -0.1 xx 7.07xx sin omega t`
`= -0.1 xx7.07 xx 0.866" "` [From equ. (1)]
`= -0.612262 = -0.61 ms^(-1)`
Kinetic energy of the block `K= (1)/(2) mv^(2)`
`=(1)/(2)xx1xx(-0.61)^(2)`
`=(1)/(2)xx0.3727 = 0.1864 = 0.19J`
Potential energy of the block `U= (1)/(2) kx^(2)`
`=(1)/(2)xx50xx(0.05)^(2) = 25xx 0.0025 = 0.0625J`
Total energy of the block at distance x= 5 cm
`E = K+U = 0.19+0.0625`
`= 0.2525 approx 0.25J`.