Question

A Body of mass 10kg is let loose from
a tower of 100m height. Calculate the
kinetie energy of the body when it is
at the bottom of the tower​

Answer 1

$\bf \red{ \underline{ \underline{Given}}}$

• Mass = 10 kg

• Height = 100 m

• g = 10 m/s²

$\bf \red { \underline{ \underline{To \: find \: out }}}$

Calculate the kinetie energy of the body when it is

at the bottom of the tower.

$\bf \red{ \underline{ \underline{Solution}}}$

$v {}^{2} = u {}^{2} + 2gh$

$\rightarrow \: (0) {}^{2} + 2 \times 10 \times 100$

$\rightarrow \: 0 + 2000$

$v {}^{2} = 2000 \: ms {}^{ - 1}$

Now,

$\purple{Kinetic \: energy \: of \: body} = \frac{1}{2} mv {}^{2}$

$\implies \: \frac{1} {2} \times 10 \times {2000} {}^{2}$

$\implies \: \frac{1}{2} \times 10 \times 4000000$

$\implies \: \frac{1} {\cancel2} \times 10 \times \cancel {4000000}$

$\implies \: 1 \times 10 \times 2000000$

= 20000000 J

Answer 2

Answer:

The potential energy at the top of tower is 9800J which is converted into kinetic energy at the bottom of tower so the kinetic energy at bottom is 9800J

Explanation: