A 9 kg mass slides to the right on a surface having a coefficient of friction 0.

Question

A 9 kg mass slides to the right on a surface having a coefficient of friction 0.37 as shown in the figure. The mass has a speed of 5 m/s when contact is made with a spring (of con- stant 107 N/m). The mass comes to rest after the spring has been compressed a distance d. The mass is then forced toward the left by the spring and continues to move in that direction beyond the unstretched position. Finally the mass comes to rest a distance D to the left of the unstretched spring. Find the distance D where the mass comes to rest. Answer in units of m

Explanation / Answer

You may have forgotten to include friction loss during spring compression KE at start from 1/2 m v^2 is 5.07 J that KE will do work against friction and convert to PE as spring is compressed so KE = 1/2 k d^2 + mu m g d 5.07 = ( 1/2*50*d^2) + (0.250*1.5 *9.81*d) 25 d^2 + 3.68 d - 5.07 = 0 d= 0.383 m so 3.68*0.383 = 2.82 J = 1.41 J is lost to sliding friction as the spring compresses it will also lose the same amount as it slides back for a total lost of 2.82 J so 5.07 - 2.82 = 2.25 J left as it leaves the spring now all of this will be lost to friction so 2.25 = 0.25*1.5*9.81*D D = 0.612 m makes sense to me

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