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Answer 1

Answer :

• the potential and the kinetic energy of a body when it falls through a distance of 15m is 250 J and 750 J respectively
• kinetic energy of the body at the ground level= 1000 J

Solution :

(a)

• mass of the body= 5 kg
• height of body from the ground = 15 m
• acceleration due to gravity =10m/s²

As the body is dropped the initial velocity of the body will be zero

• initial velocity = 0 m /s

As the initial velocity of the body is zero the initial kinetic energy will also zero

KE initial = 0

Potential energy of the body is given by the formula

$\rightarrow \mathtt{PE_i = mgh}$

$\rightarrow \mathtt{PE_i = 5 \times 10 \times 20}$

$\rightarrow \mathtt{PE_i =1000J}$

Potential energy of the body when it falls through a distance of 15 m

• height of the body from the ground = 5 m

$\rightarrow \mathtt{PE_f = mgh}$

$\rightarrow \mathtt{PE_i = 5 \times 10 \times 5}$

$\rightarrow \mathtt{PE_i =250J}$

As no external force is acting on the system the total energy of the system is conserved

By applying law of conservation of energy ,

$\rightarrow \mathtt{KE_i + PE _i = KE _ f + PE_ f }$

$\rightarrow \mathtt{0 + 1000 = KE _ f+ 250}$

$\rightarrow \mathtt{ KE _ f = 750J}$

(b)

• initial velocity of the body = 0 m/s
• acceleration due to gravity = 10 m/s²
• height of the body from the ground = 20 m

By using the third kinematic equation ,

$\star \: \boxed{ \mathtt{ {v}^{2} = {u}^{2} + 2gh}}$

here ,

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• h = height of the body from the ground

Substituting the given values in the above equation we get

$\rightarrow \mathtt{ {v}^{2} = 2gh}$

$\rightarrow \mathtt{ {v}^{2} = 2 \times 10 \times 20}$

$\rightarrow \mathtt{ {v}^{2} = 400}$

Kinetic energy of the body at ground level

$\rightarrow \mathtt{ KE = \dfrac{1}{2} m {v}^{2} }$

$\rightarrow \mathtt{ KE = \dfrac{1}{2} \times 5 \times 400 }$

$\rightarrow \mathtt{ KE = 1000J}$

Answer 2

Answer:

Answer :

the potential and the kinetic energy of a body when it falls through a distance of 15m is 250 J and 750 J respectively

kinetic energy of the body at the ground level= 1000 J

Solution :

(a)

mass of the body= 5 kg

height of body from the ground = 15 m

acceleration due to gravity =10m/s²

As the body is dropped the initial velocity of the body will be zero

initial velocity = 0 m /s

As the initial velocity of the body is zero the initial kinetic energy will also zero

KE initial = 0

Potential energy of the body is given by the formula

$\rightarrow \mathtt{PE_i = mgh}→PE$

$\rightarrow \mathtt{PE_i = 5 \times 10 \times 20}→PE$

$\rightarrow \mathtt{PE_i =1000J}→PE$

Potential energy of the body when it falls through a distance of 15 m

height of the body from the ground = 5 m

$\rightarrow \mathtt{PE_f = mgh}→PE$

$\rightarrow \mathtt{PE_i = 5 \times 10 \times 5}→PE$

$\rightarrow \mathtt{PE_i =250J}→PE$

As no external force is acting on the system the total energy of the system is conserved

By applying law of conservation of energy ,

$\rightarrow \mathtt{KE_i + PE _i = KE _ f + PE_ f }→KE$

$\rightarrow \mathtt{0 + 1000 = KE _ f+ 250}→0+1000=KE$

$\rightarrow \mathtt{ KE _ f = 750J}→KE$

=750J

(b)

initial velocity of the body = 0 m/s

acceleration due to gravity = 10 m/s²

height of the body from the ground = 20 m

By using the third kinematic equation ,

$\star \: \boxed{ \mathtt{ {v}^{2} = {u}^{2} + 2gh}}⋆$

here ,

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• h = height of the body from the ground

Substituting the given values in the above equation we get

$\rightarrow \mathtt{ {v}^{2} = 2gh}→v$

$</p><p>\rightarrow \mathtt{ {v}^{2} = 2 \times 10 \times 20}→v$

$\rightarrow \mathtt{ {v}^{2} = 400}→v$

Kinetic energy of the body at ground level

$\rightarrow \mathtt{ KE = \dfrac{1}{2} m {v}^{2} }→KE$

$\rightarrow \mathtt{ KE = \dfrac{1}{2} \times 5 \times 400 }→KE$

$\rightarrow \mathtt{ KE = 1000J}→KE=1000J$

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