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Author Question: A 25-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m). The ... (Read 60 times)

storky111

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on: Aug 1, 2020
A 25-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m). The block is pulled 10 cm to the right from its equilibrium position and released from rest. When the block has moved 2.0 cm toward its equilibrium position, its kinetic energy is 12 J. What is the change in mechanical energy caused by the frictional force on the block as it moves the 2.0 cm?

◦
−4.0 J

◦
−3.5 J

◦
−2.4 J

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−2.9 J

◦
−15 J
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Marked as best answer by storky111 on Aug 1, 2020

TheDev123

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Reply #1 on: Aug 1, 2020
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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storky111

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Reply 2 on: Aug 1, 2020
Thanks for the timely response, appreciate it

ecabral0

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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