A moving 4.0 kg block collides with a horizontal spring whose spring constant is

Question

A moving 4.0 kg block collides with a horizontal spring whose spring constant is 378 N/m (see figure). The block compresses the spring a maximum distance of 2.5 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.15. (a)What is the work done by the spring in bringing the block to rest?
(b)How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?
(c)What is the speed of the block when it hits the spring?

Explanation / Answer

   Given

                 mass =4.0 kg

                spring constant=378 N/m

                compressed distance =2.5 cm

                                                =0.025 m

           the force acting on the spring =k*x

                                                        =378*0.025

                                                        =9.45 N

                NOW   acceleration =force/mass

                                             = 9.45/4

                                             =2.3625 m/s/s .

(a) work done by the spring =force*displacment

                                           =9.45*0.025

                                           =0.23625 J

(b) friction =coefficent of friction*normal reaction

                 =0.15*4*9.8

                 =5.88 N

    the mechanical energy dissipated =5.88*0.025

                                                       =0.147 J

(c) just before hitting, K.E=P.E

                              1/2*m*v^2 =0.23625

                              1/2*4*v^2 =0.23625

                                         v =0.34 m/s .

   the velocity just before hitting the spring =0.34 m/s .

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