Question #126905
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Expert's answer
2020-07-21T12:37:03-0400

(1) According to the condition of the problem


ans. (see fig.1) a=(F2+F3)/ma=(\overrightarrow{F}_2+\overrightarrow{F}_3)/m


(2) According to the schedule the force is changed


a=dv/dta=dv/dt and dv/dt=F/mdv=(F/m)dtv=(F/m)dtdv/dt=F/m\to dv=(F/m)dt\to v=\int{(F/m)dt}


1) F=2tF=2t v1=7m/s\to v_1=7m/s


2) F=4F=4 v2=15m/s\to v_2=15m/s


3) F=3t+16F=-3t+16 v3=22.625m/s\to v_3=22.625m/s


4) F=2.5F=2.5 v4=27.625m/s\to v_4=27.625m/s Answer.


Fig.1






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