Question

A 100 grams mass What is its kinetic energy 3 seconds after the object starts falling from a height of 500 meters without any hindrance?

Answer 1

Answer :

Mass of object = 100g = 0.1g

Height of building = 500$\sf{m}$

We have to find KE of object after 3s$.$

KE of moving object is given by

• KE = 1/2 × mv²

where m denotes mass and v denotes velocity.

First we have to find velocity of object after 3s. [Equation of kinematics ☃]

After that we can calculate kinetic energy of object.

(✪) Velocity of object after 3s :

➝ v = u + gt

➝ v = 0 + (10)(3)

➝ v = 30 m/s

(✪) KE of object after 3s :

➝ KE = 1/2 × mv²

➝ KE = 1/2 × (0.1 × 30²)

➝ KE = 1/2 × 90

➝ KE = 45J

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is 45 J

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$\huge\sf\blue{Given}$

✭ Mass of body = 100 g = 0.1 kg

✭ Initial Velocity = 0 m/s

✭ Time = 3 seconds

✭ Height = 500 m

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$\huge\sf\gray{To \:Find}$

◈ The Kinetic Energy of the body?

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$\huge\sf\purple{Steps}$

So here first we have to find the velocity of the body after 3 seconds,so for that we shall use,

$\underline{\boxed{\sf v = u+gt}}$

Substituting the given values,

➳ $\sf v = u + gt$

➳ $\sf v = 0 + (10)(3)$

➳ $\sf\red{v = 30 \ m/s}$

Now that we know the velocity we shall find the kinetic energy with the help of,

$\underline{\boxed{\sf Kinetic \ Energy = \dfrac{1}{2}\times mv^2}}$

Substituting the values,

»» $\sf Kinetic \ Energy = \dfrac{1}{2}\times (0.1 \times 30^2)$

»» $\sf \dfrac{1}{2}\times 90$

»» $\sf\orange{Kinetic \ Energy = 45J}$

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