A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at x = 5 cm is 0 . 25 J . The spring constant of the spring is ______ N m - 1 .

Step-by-step solution

At first the block is pulled to a distance x 0 = 10   cm (extreme position). Potential energy of block is U i = 1 2 k x 0 2 and kinetic energy os block, K i = 0 . Now, when the block is at distance 5   cm . Potential energy is U f = 1 2 k x 0 2 2 and kinetic energy is K f = 0 . 25   J Total energy of block will be conserved, so 1 2 k x 0 2 + 0 = 1 2 k x 0 2 4 + 0 . 25 ⇒ 1 2 k x 0 2 3 4 = 1 4 ⇒ 1 2 k 3 100 = 1 ⇒ k = 200 3   N   m - 1 = 67   N   m - 1