A force of 5Newton acts on a 15kg body initially at rest. compute the work done by the force in a) the first, b) the second, and) the third seconds.and d) the instantaneous power due to the the force at the end of the third second.
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a=5/15=1/3m/s^2. _(1)
Work=F.x
a)in first second , dis placement=1/2 × a × t^2
=1/6 m
so , Work=F.x=5/6 Joule. (ANS)
b)s=1/2 at^2= 2/3 m
in 2nd second , displacement is 2/3-1/6=1/2m
so , W=5×1/2=2.5joule
c)s=1/2 at^2= 3/2 m
in 3rd second , displacement is 3/2-2/3=5/6m
so , W=5×5/6=4.17joule
(D)at the end of 3rd second ,
v=u+at
=0×(1/3)×3
=1m/s
Power=F.v
=5×1
=5J/s
I hope you are satisfied with my solution
Work=F.x
a)in first second , dis placement=1/2 × a × t^2
=1/6 m
so , Work=F.x=5/6 Joule. (ANS)
b)s=1/2 at^2= 2/3 m
in 2nd second , displacement is 2/3-1/6=1/2m
so , W=5×1/2=2.5joule
c)s=1/2 at^2= 3/2 m
in 3rd second , displacement is 3/2-2/3=5/6m
so , W=5×5/6=4.17joule
(D)at the end of 3rd second ,
v=u+at
=0×(1/3)×3
=1m/s
Power=F.v
=5×1
=5J/s
I hope you are satisfied with my solution
swaraj14:
nice explanation
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