Question #97195
A model car of mass 4kg moves in a straight line along horizontal ground. The resistance to motion of the car has magnitude 4N. The car batteries provide a constant power of 20watts. Calculate:
(i) the acceleration of the car at an instant when it's speed is 2m/s.
(I) when the car passes point A, it is moving with constant spread Vm/s. Show that V=5m/s
1
Expert's answer
2019-10-24T09:36:42-0400

Mass of car=4Kg=4 Kg

Power of batteries=20W=20 W

Resistive force=4N=4N

(i)(i) Velocity of car=2m/sec=2 m/sec

Resistive power=Fv=4×2=8W=Fv=4\times 2=8 W

Net power=20W8W=12W=20W-8W=12W

Pnet=Fnetv    Fnet=122=6NP_{net}=F_{net}v\implies F_{net}=\frac{12}{2}=6N

Fnet=ma    a=Fnetm=64=1.5m/sec2F_{net}=ma\implies a=\frac{F_{net}}{m}=\frac{6}{4}=1.5 m/sec^2

(l)(l) When speed is constant ,Resistive power== Battery power

So,P(P( batteries)=P=P (resistance)=P=20N=P=20 N ;

P=Fv    20=4×v    v=204=5m/secP=Fv\implies 20=4\times v\implies v=\frac{20}{4}=5m/sec



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