A block of mass 0.5 kg is compressed against a horizontal spring of spring constant 400 N m ⁻¹ by a distance of 0.5 m on a rough horizontal surface. Upon release, the block moves towards the natural length of the spring. If 36 % of the initial elastic potential energy is dissipated against friction, the speed of the block when the spring reaches its natural length is:

Step-by-step solution

The initial elastic potential energy stored in the spring is given by: U = 1 2 kx^2 U = 1 2 400 (0.5)^2 = 50 J It is given that 36 % of this initial energy is dissipated against friction. Therefore, the remaining mechanical energy, which converts entirely into the kinetic energy of the block at the natural length, is 100 % - 36 % = 64 % of the initial energy. Final kinetic energy K = 0.64 50 J = 32 J Equating this to the kinetic energy formula: 1 2 mv^2 = 32 1 2 0.5 v^2 = 32 0.25 v^2 = 32 v^2 = 128 v = 128 = 8 2 m