The potential energy of a body of mass 0.5 kg increases by 100 J when it is taken to the top of a tower from ground. If force of gravity on 1 kg is 10 N. what is the height of the tower ?

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Solution

Force of gravity on 1kg is 10 N
F = mg
10 = 1 * g
g = 10 $bevelled fraction numerator m over denominator s e c squared end fraction$

Change in Potential energy = mg * ( h2 - h1 )
= mg * ( h - 0 )
= mgh

100 = mgh
h = $fraction numerator 100 over denominator m g end fraction$
h = $fraction numerator 100 over denominator 0.5 cross times 10 end fraction$
h = 20 m

Height of the tower is 20 m

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The potential energy of a body of mass 0.5 kg increases by 100 J when it is taken to the top of a tower from ground. If force of gravity on 1 kg is 10 N. what is the height of the tower ?

Force of gravity on 1kg is 10 N F = mg 10 = 1 * g g = 10 Change in Potential energy = mg * ( h2 - h1 ) = mg * ( h - 0 ) = mgh 100 = mgh h = h = h = 20 m Height of the tower is 20 m