A force of 5N is applied on a 20kg mass at rest.wirk done in the third second
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Answered by
32
we know that sn=u+a/2 (2n-1)
by applying this we get
sn=0+5/20 (2×3-1)
sn=0.25 (5)
sn=1.25
here,
work done=fs then
w.d=5×1.25=6.25answer
by applying this we get
sn=0+5/20 (2×3-1)
sn=0.25 (5)
sn=1.25
here,
work done=fs then
w.d=5×1.25=6.25answer
Answered by
32
But the answer option is in 25/8 J
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