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.

A block whose mass is 1kg is fastened to a spring. The spring has a sp

1.	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.
Answer» Solution :`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`.