Physics, asked by varunpoonia21, 3 months ago

an object of mass 40 kg is raised to a height of 5m above the ground what is its potential energy if the object is allowed to fall find its kinetic energy when it is half way down​

Answers

Answered by MystícPhoeníx
236

Answer:

  • 1960 J & 980 J is the required energy.

Explanation:

Given:-

  • Mass of Object ,m = 40kg
  • height ,h = 5 m

To Find:-

  • Potential Energy
  • Kinetic Energy when it is half way down

Solution:-

As we know that the acceleration due to gravity on earth surface is 9.8m/.

Potential Energy is defined as the energy possess by body due to its position.

PE = mgh

where,

  • PE is Potential Energy
  • m is the mass
  • g is the acceleration due to gravity
  • h is the height

Substitute the value we get

→ PE = 40×9.8×5

→ PE = 200 × 9.8

→ PE = 1960 J

  • Hence, Potential Energy at maximum height is 1960 Joules.

As we know that Law of Conservation of energy states that energy can neither be created or nor be destroyed . It can be transferred from one state to another. Potential Energy convert into Kinetic. As the object is allowed to fall in its half way .

Therefore, new height of object = 5/2 = 2.5 m

KE = mgh

Substitute the value we get

→ KE = 40×9.8×2.5

→ KE = 4×98 × 2.5

→ KE = 980 J

  • Hence, the kinetic energy of the body at half way of its position is 980 Joules.
Answered by Anonymous
111

Answer:

Given :-

  • Mass = 40 kg
  • Height = 5 m

To Find :-

  • PE
  • object is allowed to fall find its kinetic energy when it is half way down

Solution :-

As we know that

 \huge \fbox{PE = Mgh}

PE is the Potential Energy

M is the Mass

G is the Acceleration due to gravity

H is the Height

 \sf \: PE = 40 \times 9.8 \times 5

  • Taking g as 9.8 m/

 \sf \: PE = 200 \times 9.8

 \frak \red{PE= 1960 J}

Now,

When object fall half down

Height will be divided by 2

5/2 = 2.5 m

 \huge \fbox{KE = mgh}

KE is the Kinetic energy

M is the Mass

G is the Acceleration due to gravity

H is the Height

 \sf \: KE = 40 \times 2.5 \times 9.8

 \sf \: KE \:  = 4 \times 98 \times 2.5

  \frak \red{KE \:  = 980 \: joules{}}

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