Question

A 75 kg box is dropped from the top of a tower. The height of the tower is 35m. Calculate-
1.The initial potential energy of the box.
2.Its potential energy 15m above the ground.
3.The maximum value of its kinetic energy. and
4.Its kinetic energy 20m below the top of the tower.​

Answer 1

Answer:

potential energy =mgh

=75*10*35

=26250N

h=35-15=20

potential energy=mgh

=75*10*20

=15000

Answer 2

Answer:

Answer:

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:

$\textsf{Mass of the box = 75 kg}$

$\textsf{Height of the tower = 35 m}$

$\dag$To find:

$\textsf{1. The initial energy of the box}$

$\textsf{2. Its potential energy 15 m above the ground}$

$\textsf{3. The maximum value of its kinetic energy}</p><p></p><p>[tex]\textsf{4. Its kinetic energy 20 m below the top of the tower}$

$\dag$Taken:

Formula for number 1 :

$\boxed{PI=mgh}$

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 :

$\boxed{Pe=mghi}$

Where,

Pe = Potential energy

m = Mass of the box

g = Acceleration due to gravity

hi = Height above the ground

Formula for number 3 :

$\boxed{Ke=\frac{1}{2}mv²}$

Where,

Ke = Kinetic energy

m = Mass of the box

v = Velocity of the box

Formula for number 4 :

$\boxed{Ke=\frac{1}{2}mva²}$

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:

$\boxed{ = \sqrt{2gh} }$

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 :

$\mapsto{\sqrt{2×9.8×35}}$

$\mapsto{\sqrt{19.6×25}}$

$\mapsto{\sqrt{686}}$

$\mapsto{26.195\:m/s(approx)}$

Now , at 15 m :

$\mapsto{\sqrt{2×9.8×15}}$

$\mapsto{\sqrt{19.6×15}}$

$\mapsto{\sqrt{294}}$

$\mapsto{17.146\:m/s(approx)}$

$\dag$Solution:

Number 1 :

$Taken,PI=mgh$

$\Rightarrow{PI=75kg×9.8m/s×35m}$

$\Rightarrow{PI=735×35}$

$\Rightarrow{PI=25725J}$

So , Initial potential energy is 25725 joules.

___________________________________

Number 2 :

$Taken,Pe=mghi$

$\Rightarrow{Pe=75kg×9.8m/s×15m}$

$\Rightarrow{Pe=735×15m}$

$\Rightarrow{Pe=11025}$

So , the potential energy at 15 m above the ground is 11025 joules .

___________________________________

Number 3 :

$Taken,Ke=\frac{1}{2}mv²$

$\Rightarrow{Ke=\frac{1}{2}×75kg×26.195m/s²}$

$\Rightarrow{Ke=37.5kg×26.195m/s²}$

$\Rightarrow{Ke=37.5kg×(26.195×26.195)}$

$\Rightarrow{Ke=37.5kg×686(approx)}$

$\Rightarrow{Ke=25725J}$

So , the maximum kinetic energy is 25725 joules .

___________________________________

Number 4 :

$Taken,Ke=\frac{1}{2}mva²$

$\Rightarrow{Ke=\frac{1}{2}×75kg×17.146m/s²}$

$\Rightarrow{Ke=37.5kg×17.146m/s²}$

$\Rightarrow{Ke=37.5kg×(17.146×17.146)}$

$\Rightarrow{Ke=37.5×294(approx)}$

$\Rightarrow{Ke=11025J}$

So , Kinetic energy below the 20 m of the tower is 11025 joules .

Extra information:

First equation of motion:

$Vi=Vf-at$

Where,

Vi = Initial velocity

Vf = Final velocity

a = Acceleration

t = Time taken

Second equation of motion:

$v=ut+\frac{1}{2}at²$

Where,

v = Final velocity

u = Initial velocity

t = Time taken

a = Acceleration

Third equation of motion:

$2as=v²+u²$

Where,

a = Acceleration

s = Displacement

v² = Final velocity

u² = Initial velocity