Question

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

Answer 1

Answer:-

$\implies$1115.2 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 :-

$\maltese$As the block falls , the change in potential energy will be converted into kinetic energy

$\leadsto$Potential Energy at rest at height,$h_2$ = $\sf mgh_2$

$\leadsto$Potential Energy when fell at height,$h_1$ = $\sf mgh_1$

$\leadsto$Kinetic energy= Change in Potential energy = = $\sf mgh_2-mgh_1$

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

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

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

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

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