Question

A moving 5.00 kg block collides with a horizontal spring whose spring constant is 468 N/m. The block compresses the spring a maximum distance of 2.50 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.320. What is the work done by the spring in bringing the block to rest? -1.46x10-1 J Computer's answer now shown above. Submission not graded. Use more digits. Tries 0/12 Previous Tries How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? 3.92x10-1 Computer's answer now shown above. Tries 1/12 Previous Tries What is the speed of the block when it hits the spring? 4.64x101 m/s Computer's answer now shown above. Tries 1/12 Previous Tries

What is the conceptual and mathematical methods for solving this problem?

Answer 1

Since the spring is compressed 2.50 cm, the work done by the spring will be equal to change in potential energy, which is W-K Corg-a;?) = -1 x kg Wc-4x468 x12.50 X10-2) : W=-1468X10-4 Nm - =-146X10 Nm W = -1.46 10- J Since the friction force acting on the block is ak F = Mmg a I work done by friction, which is equal to dissipated energy is . We = mmg.d - 50-320 X 59.8% 2.50X10-2 . Wp = 3.92 xló 'J By work-energy theorem, the change in kinetic energy is equal to work done by the spring & friction i K=my? 1.46 X10-1 +3.92 X10-! mu? 0.14610-392 v2= 2X0.538 1² 0.2152 ; V= 0.4639 mis 3v=4.64 X10 mis