The potential energy of a body of mass $0 \cdot 5 k g$ increases by $100 J$ when it is taken to the top of a tower from ground. If force of gravity on $1 k g$ is $10 N$ , what is the height of the tower?

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Solution

Given that,

Mass of body $M = 0.5 k g$

Potential energy $P . E = 100 J$

Force of gravity on $1 k g$ is $10 N$

Now,

$F = m g$

$10 = 1 \times g$

$g = 10 m / s^{2}$

Now, change in Potential energy

$= m g (h_{2} - h_{1})$

$= m g (h - 0)$

$= m g h$

Now, the height is

$100 = m g h$

$h = \frac{100}{0.5 \times 10}$

$h = 20 m$

Hence, the 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?

Given that, Mass of body M=0.5kg Potential energy P.E=100J Force of gravity on 1 kg is 10 N Now, F=mg 10=1×g g=10m/s2 Now, change in Potential energy =mg(h2−h1) =mg(h−0) =mgh Now, the height is 100=mgh h=1000.5×10 h=20m Hence, the height of the tower is 20 m

The potential energy of a body of mass $0 \cdot 5 k g$ increases by $100 J$ when it is taken to the top of a tower from ground. If force of gravity on $1 k g$ is $10 N$ , what is the height of the tower?