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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

Sagot :

Answer:-

[tex]\implies[/tex]1115.2 Joules

Given :-

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

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

Solution :-

[tex]\maltese[/tex]As the block falls , the change in potential energy will be converted into kinetic energy

[tex]\leadsto[/tex]Potential Energy at rest at height,[tex]h_2[/tex] = [tex]\sf mgh_2[/tex]

[tex]\leadsto[/tex]Potential Energy when fell at height,[tex]h_1[/tex] = [tex]\sf mgh_1[/tex]

[tex]\\[/tex]

[tex]\leadsto[/tex]Kinetic energy= Change in Potential energy = = [tex]\sf mgh_2-mgh_1[/tex]

[tex]\begin{gathered}\implies\quad \sf mg(h_2-h_1) \\\end{gathered} [/tex]

[tex]\begin{gathered}\implies\quad \sf mg(53.0-21.3) \\\end{gathered} [/tex]

[tex]\begin{gathered}\implies\quad \sf 3.59 \times 9.8\times (53.0-21.3) \\\end{gathered} [/tex]

[tex]\begin{gathered}\implies\quad \sf 3.59 \times 9.8\times 31.7\\\end{gathered} [/tex]

[tex]\begin{gathered}\implies\quad \boxed{\sf{1115.2\: Joules}}\\\end{gathered} [/tex]