Question #297577

A 5000 KG TRUCK IS RUNNING AT 72 KPH.

a. ) IF IT IS TO BE STOPPED BY A CONSTANT RETARDING FORCE IN 10 S. WHAT IS THE DECELERATION ?

b ) WHAT IS THE RETARDING FORCE ?


1
Expert's answer
2022-02-15T16:14:51-0500

Given:

m=5000kgm=\rm 5000\: kg

vi=20m/sv_i=\rm 20\: m/s

vf=0m/sv_f=\rm 0\: m/s

Δt=10s\Delta t=\rm 10\: s


(a) the deceleration of the truck


a=ΔvΔt=2010=2m/s2a=\frac{\Delta v}{\Delta t}=\frac{-20}{10}=\rm -2\: m/s^2

(b) the retarding force


F=ma=50002=10000NF=ma=5000*2=10000\:\rm N

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