A horizontal 40 cm spring with a spring constant of 250 N/m is attached to a wal

Question

A horizontal 40 cm spring with a spring constant of 250 N/m is attached to a wall and a 4 kg mass as shown in the figure at right. The 4 kg mass slides on a horizontal surface with a coefficient of kinetic friction of 0.15. The spring is stretched to a length of 50 cm and then released. After several oscillations of the block back and forth on the table, the entire system comes to a rest with the length of the spring returned again to 40 cm. What was the total distance travelled by the block (in m) during this process? (Hint: I am not asking for the difference between the initial and final positions of the block; rather, I want to know what you would calculate if you added up the total amplitude of each oscillation of the block).

Explanation / Answer

Initial Spring energy provided = Kx2/2

Change in KE = 0 [Since body comes back to rest].

Now, Final Spring energy = 0 . [Returns to initial position]

Work done by friction = Initial energy lost = Kx2/2

f * Distance = Kx2/2

mg * Distance = Kx2/2

Distance = 5.314 m [g = 9.8]

= 5.2 m [g = 10]

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