Find $m$ , if $v = 3, g = 10, h = 5$ and $E = 109$ .

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Solution

The correct option is B $2$
Given, $E = \frac{1}{2} m v^{2} + m g h$
$\Rightarrow E = \frac{1}{2} (m v^{2} + 2 m g h)$
$\Rightarrow 2 E = m v^{2} + 2 m g h$
$\Rightarrow 2 E = m (v^{2} + 2 g h)$
$\Rightarrow m = \frac{2 E}{v^{2} + 2 g h}$

Substituting $v = 3, g = 10, h = 5, E = 109$ , we get
$m = \frac{2 \times 109}{3^{2} + 2 \times 10 \times 5} = \frac{2 \times 109}{9 + 100} = \frac{2 \times 109}{109} = 2$

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Q. The total energy E possessed by a body of mass 'm', moving with a velocity 'v' at a height 'h' is given by :

E = \frac{1}{2} m v^{2} + m g h .

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and

E = 109

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respectively are arranged as shown in figure.
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Find m, if v=3,g=10,h=5 and E=109.

The correct option is B 2Given, E=12mv2+mgh⇒E=12(mv2+2mgh)⇒2E=mv2+2mgh⇒2E=m(v2+2gh)⇒m=2Ev2+2ghSubstituting v=3,g=10,h=5,E=109, we get m=2×10932+2×10×5=2×1099+100=2×109109=2

Find $m$ , if $v = 3, g = 10, h = 5$ and $E = 109$ .