Question #285670

How much work is required to accelerate a 1500kg car from 15 m/s to 40 m/s?What is the average net force acting on the car if it covers a distance of 275m? 


1
Expert's answer
2022-01-10T09:09:11-0500

Given:

v1=15m/sv_1=\rm 15\: m/s

v2=40m/sv_2=\rm 40\: m/s

m=1500kgm=\rm 1500\: kg

d=275md=\rm 275\: m


The energy-work theorem says

mv222mv122=W=Fd\frac{mv_2^2}{2}-\frac{mv_1^2}{2}=W=Fd

(a) the work done

W=mv222mv122=15002(402152)=1031250JW=\frac{mv_2^2}{2}-\frac{mv_1^2}{2}\\=\frac{1500}{2}(40^2-15^2)=1\:031\:250\: \rm J

(b) the average force

F=Wd=1031250275=3750NF=\frac{W}{d}=\frac{1031250}{275}=3\:750\: \rm N


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