A force of 20 N acts on 16 kg block kept on a smooth horizontal surface for 6 seconds and then removed. find the distance covered by it in 10 s
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here F=ma
20=16a
a=5/4m/sec^2
u=0 t=6sec
S= ut+1/2 at^2
S= 0+1/2×5/4×6×6
S= 45/2m
from here v=15/2m/sec
this v is initial for next 4 secs then
u=15/2 t=4sec
S= 15/2×4= 30m
total distance = 30+45/2=105/2m
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