A force acts from 8sec on …
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Ankit Kumar 8 years, 11 months ago
applied force = opposite force causing the body to be stop i.e., terminal velocity v=0m/s
s=ut+at2
0.8=ux5+ax52
0.8=5u+25a/2
1.6=10u+25a:.........................eq.1
now acceleration a=
u+5a=0:.....................................eq.2
on solving eq.1 and eq.2 we get
a= -0.064m
force=ma
so required force =10x0.064
=0.64N
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