A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m^(-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 1 kg is fastened to a spring. The spring has a s

1.	A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m^(-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 :The block EXECUTES SHM, its angular frequency, as given by Eq. (14.14b), is `omega=sqrt(k/m)` `=sqrt((50Nm^(-1))/(1kg))` `=7.07rads^(-1)` Its displacement at any time t is then given by, x(t) = 0.1 cos (7.07t) Therefore, when the particle is 5 cm away from the MEAN position, we have 0.05 = 0.1 cos (7.07t) Or cos (7.07t) = 0.5 and hence `sin(7.07t)=(sqrt3)/2=0.866` Then, the velocity of the block at x = 5 cm is `=0.1xx7.07xx0.866ms^(-1)` `=0.61ms^(-1)` Hence the K.E. of the block, `=1/2mv^(2)` `=1/2[1kgxx(0.6123ms^(-1))^(2)]` `=0.19J` The P.E. of the block, `=1/2kx^(2)` `=1/2(50Nm^(-1)xx0.05mxx0.05m)` `=0.0625J` The TOTAL energy of the block at x = 5 cm, = K.E. + P.E. = 0.25 J we also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system, `=1/2(50Nm^(-1)xx0.1mxx0.1m)` =0.25J which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of CONSERVATION of energy.