A ball of mass 0.5 kg is dropped from a height. Find the work done by its weight in one second after the ball is dropped
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Answered by
14
f=mg=0.5×9.8=4.9kg/ms^2
because acceleration is =9.8m/s^2
therefore velocity in one second=9.8/1=9.8m/s^2
Displacement in one second=9.8/1=9.8m/s^2
therefore work done=4.9×9.8=48.02joul
because acceleration is =9.8m/s^2
therefore velocity in one second=9.8/1=9.8m/s^2
Displacement in one second=9.8/1=9.8m/s^2
therefore work done=4.9×9.8=48.02joul
Answered by
9
Given:-
Mass = 0.5kg
Time = 1 sec
Since
F = mg
F = 0.5*9.8
= 4.9N
Displacement in 1 sec = 9.8m
As acceleration will be 9.8ms^-2
We know
W = F* D
= 4.9*9.8
=48.02J
Mass = 0.5kg
Time = 1 sec
Since
F = mg
F = 0.5*9.8
= 4.9N
Displacement in 1 sec = 9.8m
As acceleration will be 9.8ms^-2
We know
W = F* D
= 4.9*9.8
=48.02J
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