A box of mass 10 , kg is placed at point A on | vedclass

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(A) The mass of the box is $m = 10 \, kg$.The gravitational force (weight) acting on the box is $F = mg = 10 \, kg 	imes 10 \, m s^{-2} = 100 \, N$,w

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$0$

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(A) The mass of the box is $m = 10 \, kg$.The gravitational force (weight) acting on the box is $F = mg = 10 \, kg 	imes 10 \, m s^{-2} = 100 \, N$,which acts vertically downwards.The displacement $S = 2 \, m$ is along the horizontal surface.The angle $	heta$ between the gravitational force (downward) and the displacement (horizontal) is $90^{\circ}$.The work done $W$ is given by the formula $W = FS \cos 	heta$.Substituting the values: $W = 100 	imes 2 	imes \cos 90^{\circ}$.Since $\cos 90^{\circ} = 0$,the work done $W = 100 	imes 2 	imes 0 = 0 \, J$.Therefore,the work done by the gravitational force is $0 \, J$ because the force of gravity is perpendicular to the direction of displacement.

Height of object $(m)$	$PE$ $(J)$	$KE$ $(J)$	Mechanical Energy $(J)$
$10$	...	...	...
$8$	...	...	...
$5$	...	...	...
Just above ground	...	...	...

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