Question

A 28 kg block on a horizontal surface is attached to a horizontal spring of spring constant k = 2.9 kN/m. The block is pulled to the right so that the spring is stretched 8.4 cm beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of 37 N. (a) What is the kinetic energy of the block when it has moved 2.7 cm from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

Answer 1

initial potential energy stored when spring stretched = 1/2*k*x^2 = 8.82 J

a) kinetic energy of block after it moved 2.7 cm from its point of release = 0.5 * 28 * v^2 where v is its instantaneous velocity at that point of time

potential energy of spring at this point of time = 0.5 * k *(8.4 - 2.7)^2 = 4.71 J

energy dissipated in friction = 37*.027 = 0.999 J

according to law of conservation of energy

energy dissipated in friction + potential energy of spring at this point of time +  kinetic energy of block after it moved 2.7 cm from its point of release = initial potential energy stored when spring stretched

=> kinetic energy of block =  8.82 - 4.71 - 0.999 = 3.111 J

b) potential energy of spring when spring is relaxed = 0

energy dissipated in friction = 3.108

kinetic energy of block when it first crosses the relaxed point +  potential energy of spring when spring is relaxed + energy dissipated in friction = initial potential energy stored when spring stretched

=> kinetic energy of block when it first crosses the relaxed point = 8.82 - 3.108 = 5.712 J

c) maximum kinetic energy = 5.712 Joule