A ball is released from the top of a tower the ratio of work done by force of gravity in first,secind and third second is
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Hi mate here is your answer,
For t=1
S=ut+1/2gt^2 =1/2gt^2. (since u=0 m/s)
S=1/2gt^2=1/2×g=g/2
Similarly for t=2
S=1/2×g×4=2g
And, For t=3
S=9g/2
So, the distance traveled in 1st, 2nd and 3rd second are =g/2 ,( 2g-g/2),(9g/2-2g)
=g/2, 3g/2, 5g/2
So, the work done will be =Force × displacement
=mg×g/2, mg×3g/2, mg×5g/2
=mg^2/2, mg^2 3/2, mg^2 5/2
In ratio , we can cancel out mg^2 1/2
So, answer is =1:3:5
For t=1
S=ut+1/2gt^2 =1/2gt^2. (since u=0 m/s)
S=1/2gt^2=1/2×g=g/2
Similarly for t=2
S=1/2×g×4=2g
And, For t=3
S=9g/2
So, the distance traveled in 1st, 2nd and 3rd second are =g/2 ,( 2g-g/2),(9g/2-2g)
=g/2, 3g/2, 5g/2
So, the work done will be =Force × displacement
=mg×g/2, mg×3g/2, mg×5g/2
=mg^2/2, mg^2 3/2, mg^2 5/2
In ratio , we can cancel out mg^2 1/2
So, answer is =1:3:5
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