A small 2 kg mass moved slowly from the surface of earth to a height of 6.4 × 106m above theearth. Find the work done [in mega joule].

# A small 2 kg mass moved slowly from the surface of earth to a height of  above theearth. Find the work done [in mega joule].

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### Solution:

${\mathrm{W}}_{\mathrm{ext}}=\mathrm{\Delta U}={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{1}=\left(\frac{-\mathrm{GM}}{\left(\mathrm{R}+\mathrm{h}\right)}\right)-\left(\frac{-\mathrm{GM}}{\mathrm{R}}\right)$

$\mathrm{\Delta U}=\frac{2×10×6.4×{10}^{6}}{2}$

$⇒\mathrm{\Delta U}=6.4×{10}^{7}\text{Joule}⇒\mathrm{W}=\mathrm{\Delta U}=64\mathrm{MJ}$

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