A moving 4.30 kq block collides with a horizontal spring whose spring constant i

Question

A moving 4.30 kq block collides with a horizontal spring whose spring constant is 364 N/m. The block compresses the spring a maximum distance of 14.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.420. What is the work done by the spring in bringing the block to rest? Consider DeltaK+DeltaU=DeltaE. What quantities in the problem correspond to each of the variables in the equation? How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? What is the speed of the block when it hits the spring?

Explanation / Answer

work done by spring = -0.5 kx^2

                              = - 0.5*364*0.14*0.14

                          = -3.5672 J

Energy dissipated by friction = uMgx

                     = 0.42*4.3*9.8*0.14

                    = 2.4778 J

total initial KE = Energy dissipated by friction - work done by spring

                       =2.4778 +3.5672 J = 6.045 J

0.5 M v^2 = 6.045

v^2 = 6.045/[0.5*4.3]

        = 2.8116

v = 1.677 m/s

speed when it hits the spring is 1.677 m/s

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