A block of mass m is attached to one end of a horizontal ideal spring of spring constant k , while the other end is fixed. The block is pushed to compress the spring by a distance A and then released from rest on a smooth horizontal surface. At a certain instant during its motion, the kinetic energy of the block is twice the elastic potential energy of the spring. The compression of the spring and the speed of the bl

Step-by-step solution

Total energy of the system is E = 1 2 kA^2 . Let x be the compression of the spring and v be the speed of the block at the given instant. The elastic potential energy is U = 1 2 kx^2 and the kinetic energy is K = 1 2 mv^2 . Given that K = 2U , by conservation of mechanical energy: E = K + U = 3U 1 2 kA^2 = 3 ( 1 2 kx^2 ) x^2 = A^2 3 x = A 3 Now, substituting x into K = 2U : 1 2 mv^2 = 2 ( 1 2 k ( A 3 )^2 ) 1 2 mv^2 = kA^2 3 v = A 2k 3m Answer: A 3 , A 2k 3m