A 75 kg box is dropped from the top of a tower. The height of the tower is 35 m. Calculate (i) the initial
potential energy of the box, (ii) its potential energy 15 m above the ground, (iii) the maximum value of
its kinetic energy and (iv) its kinetic energy 20 m below the top of the tower. (g = 9.8 m/s2)
(g = 9.8 m/s2)
Answers
1. Initial potential energy of the box = 25725 J
2. Potential energy 15 m above the ground = 11025 J
3. Maximum value of its kinetic energy = 25725 J
4. Kinetic energy 20 m below the top of the tower = 11025 J
\dag† Given:
\textsf{Mass of the box = 75 kg}Mass of the box = 75 kg
\textsf{Height of the tower = 35 m}Height of the tower = 35 m
\dag† To find:
\textsf{1. The initial energy of the box}1. The initial energy of the box
\textsf{2. Its potential energy 15 m above the ground}2. Its potential energy 15 m above the ground
\textsf{3. The maximum value of its kinetic energy} < /p > < p > < /p > < p > [tex]\textsf{4. Its kinetic energy 20 m below the top of the tower}3. The maximum value of its kinetic energy</p><p></p><p>[tex]4. Its kinetic energy 20 m below the top of the tower
\dag† Taken:
Formula for number 1 :
\boxed{PI=mgh}
PI=mgh
Where,
PI = Initial potential energy
m = Mass of the box
g = Acceleration due to gravity
h = Heigth of the tower
( Acceleration due to gravity is 9.8 m/s )
Formula for number 2 :
\boxed{Pe=mghi}
Pe=mghi
Where,
Pe = Potential energy
m = Mass of the box
g = Acceleration due to gravity
hi = Height above the ground
Formula for number 3 :
\boxed{Ke=\frac{1}{2}mv²}
Ke=
2
1
mv²
Where,
Ke = Kinetic energy
m = Mass of the box
v = Velocity of the box
Formula for number 4 :
\boxed{Ke=\frac{1}{2}mva²}
Ke=
2
1
mva²
Where,
Ke = Kinetic energy
m = Mass of the box
va = Velocity at 20 m below the tower
\dag† Concept:
See , to find the kinetic energy first we have to find the velocity of the box by this equation:
\boxed{ = \sqrt{2gh} }
=
2gh
Where,
g = Acceleration due to gravity
h = Height of the tower at where we have to find the kinetic energy
So , now first find the velocity of the box at the height of 35 m and 35 - 20 = 15 m.
First at 35 m :
\mapsto{\sqrt{2×9.8×35}}↦
2×9.8×35
\mapsto{\sqrt{19.6×25}}↦
19.6×25
\mapsto{\sqrt{686}}↦
686
\mapsto{26.195\:m/s(approx)}↦26.195m/s(approx)
Now , at 15 m :
\mapsto{\sqrt{2×9.8×15}}↦
2×9.8×15
\mapsto{\sqrt{19.6×15}}↦
19.6×15
\mapsto{\sqrt{294}}↦
294
\mapsto{17.146\:m/s(approx)}↦17.146m/s(approx)
\dag† Solution:
Number 1 :
Taken,PI=mghTaken,PI=mgh
\Rightarrow{PI=75kg×9.8m/s×35m}⇒PI=75kg×9.8m/s×35m
\Rightarrow{PI=735×35}⇒PI=735×35
\Rightarrow{PI=25725J}⇒PI=25725J
So , Initial potential energy is 25725 joules.
___________________________________
Number 2 :
Taken,Pe=mghiTaken,Pe=mghi
\Rightarrow{Pe=75kg×9.8m/s×15m}⇒Pe=75kg×9.8m/s×15m
\Rightarrow{Pe=735×15m}⇒Pe=735×15m
\Rightarrow{Pe=11025}⇒Pe=11025
So , the potential energy at 15 m above the ground is 11025 joules .
___________________________________
Number 3 :
Taken,Ke=\frac{1}{2}mv²Taken,Ke=
2
1
mv²
\Rightarrow{Ke=\frac{1}{2}×75kg×26.195m/s²}⇒Ke=
2
1
×75kg×26.195m/s²
\Rightarrow{Ke=37.5kg×26.195m/s²}⇒Ke=37.5kg×26.195m/s²
\Rightarrow{Ke=37.5kg×(26.195×26.195)}⇒Ke=37.5kg×(26.195×26.195)
\Rightarrow{Ke=37.5kg×686(approx)}⇒Ke=37.5kg×686(approx)
\Rightarrow{Ke=25725J}⇒Ke=25725J
So , the maximum kinetic energy is 25725 joules .
___________________________________
Number 4 :
Taken,Ke=\frac{1}{2}mva²Taken,Ke=
2
1
mva²
\Rightarrow{Ke=\frac{1}{2}×75kg×17.146m/s²}⇒Ke=
2
1
×75kg×17.146m/s²
\Rightarrow{Ke=37.5kg×17.146m/s²}⇒Ke=37.5kg×17.146m/s²
\Rightarrow{Ke=37.5kg×(17.146×17.146)}⇒Ke=37.5kg×(17.146×17.146)
\Rightarrow{Ke=37.5×294(approx)}⇒Ke=37.5×294(approx)
\Rightarrow{Ke=11025J}⇒Ke=11025J
So , Kinetic energy below the 20 m of the tower is 11025 joules .
Extra information:
First equation of motion:
Vi=Vf-atVi=Vf−at
Where,
Vi = Initial velocity
Vf = Final velocity
a = Acceleration
t = Time taken
Second equation of motion:
v=ut+\frac{1}{2}at²v=ut+
2
1
at²
Where,
v = Final velocity
u = Initial velocity
t = Time taken
a = Acceleration
Third equation of motion:
2as=v²+u²2as=v²+u²
Where,
a = Acceleration
s = Displacement
v² = Final velocity
u² = Initial velocity
Answer:
अहिंसा का सिद्धान्त जैन धर्म की मुख्य षिक्षा है। जैन धर्म में पशु-पक्षी तथा पेड़-पौधे तक की हत्या न करने का अनुरोध किया गया है। अहिंसा की शिक्षा से ही समस्त देश में दया को ही धर्म प्रधान अंग माना जाता है। ... पंचशील सिद्धान्त के प्रर्वतक एवं जैन धर्म के चैबीसवंे तीर्थंकर महावीर स्वामी अहिंसा के मूर्तिमान प्रतीक थे।