a particle of mass 0. 5kg is kept at rest. a force of 2.0N acts on it for 5.0s.find the distance moved by the partical in(a) these 5.0s,and (b) the next 5.0s
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F=ma
a=F/m=2/0.5=4 m/s^2
u=0 m/s
S=0.5at^2
S=0.5*4*25
S=12.5*4=50 m
(a.)distance moved by particle in 5 seconds is 50 m
similarly
S=0.5*4*10*10=200 m
then the distance moved in the next 5 seconds is
(b.)200-50 = 150 m
a=F/m=2/0.5=4 m/s^2
u=0 m/s
S=0.5at^2
S=0.5*4*25
S=12.5*4=50 m
(a.)distance moved by particle in 5 seconds is 50 m
similarly
S=0.5*4*10*10=200 m
then the distance moved in the next 5 seconds is
(b.)200-50 = 150 m
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