If a body of mass 200, g falls from a height | vedclass

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(B) The mass of the body is $m = 200\, g = 0.2\, kg$.The height from which the body falls is $h = 200\, m$.The acceleration due to gravity is $g = 10\

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) The mass of the body is $m = 200\, g = 0.2\, kg$.The height from which the body falls is $h = 200\, m$.The acceleration due to gravity is $g = 10\, m/s^2$.The potential energy $(P.E.)$ at the initial height is given by $U = mgh$.At the point of contact with the earth's surface,the final height is $0$,so the final potential energy is $0$.The decrease in $P.E.$ is $\Delta U = U_{initial} - U_{final} = mgh - 0 = mgh$.Substituting the values: $\Delta U = 0.2\, kg 	imes 10\, m/s^2 	imes 200\, m = 400\, J$.Therefore,the decrease in $P.E.$ is $400\, J$.

If a body of mass $200\, g$ falls from a height $200\, m$ and its total $P.E.$ is converted into $K.E.$ at the point of contact of the body with the earth's surface,then what is the decrease in $P.E.$ of the body at the contact? $(g = 10\, m/s^2)$ ............ $J$

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