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Author Question: The two masses in the figure are released from rest. After the 3.0-kg mass has fallen 1.5 m, it is ... (Read 223 times)

Mr3Hunna

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on: Aug 1, 2020

Question 1

The speed of a 4.0-kg object is given by v = (2t) m/s, where t is in s. At what rate is the resultant force on this object doing work at t = 1 s?

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48 W

◦
40 W

◦
32 W

◦
56 W

◦
16 W


Question 2

The two masses in the figure are released from rest. After the 3.0-kg mass has fallen 1.5 m, it is moving with a speed of 3.8 m/s. What is the change in mechanical energy done on the system during this time interval by the frictional force on the 2.0 kg mass?



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−12 J

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−17 J

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−20 J

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−8.0 J

◦
−28 J
Physics for Scientists and Engineers ¦ Serway, Jewett ¦ 9th Edition
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Marked as best answer by Mr3Hunna on Aug 1, 2020

jamesnevil303

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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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Mapo

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Reply #2 on: Nov 17, 2023
Thank You


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