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A 3.59 kg block is released from rest 53.0 m above the ground. When it has fallen 21.3 m, what is its kinetic energy

1 Answer

13 votes

Answer:-


\implies1115.2 Joules

Given :-


\red{\leadsto}\:
\textsf{Height\: at \:rest\:}
\sf,h_2= 53.0m


\green{\leadsto}\:
\textsf{Height\: to\:which\;block\:}
\sf,h_1= 21.3

Solution :-


\malteseAs the block falls , the change in potential energy will be converted into kinetic energy


\leadstoPotential Energy at rest at height,
h_2 =
\sf mgh_2


\leadstoPotential Energy when fell at height,
h_1 =
\sf mgh_1


\\


\leadstoKinetic energy= Change in Potential energy = =
\sf mgh_2-mgh_1


\begin{gathered}\implies\quad \sf mg(h_2-h_1) \\\end{gathered}


\begin{gathered}\implies\quad \sf mg(53.0-21.3) \\\end{gathered}


\begin{gathered}\implies\quad \sf 3.59 * 9.8* (53.0-21.3) \\\end{gathered}


\begin{gathered}\implies\quad \sf 3.59 * 9.8* 31.7\\\end{gathered}


\begin{gathered}\implies\quad \boxed{\sf{1115.2\: Joules}}\\\end{gathered}

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