Question

A body of mass 15 kg falls from a height of 160 cm. Calculate: i) velocity of the body ii) K.E of the body when it just reaches the ground. (g = 9.8 SI unit)

Answer 1

$\orange{\bigstar}$  Given  $\green{\bigstar}$

A body of mass 15 kg falls from a height of 160 cm

$\orange{\bigstar}$  To Find  $\green{\bigstar}$

i. Velocity of the body

ii. K.E of the body when it just reaches the ground

$\orange{\bigstar}$  Concept Used  $\green{\bigstar}$

3rd Equation of motion

$\bf \pink{\bigstar\ \; v^2-u^2=2as}$

where ,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes height

K.E of the body

$\bf \green{\bigstar\ \; K.E =\dfrac{1}{2}mv^2}$

where ,

• K.E denotes kinetic energy
• m denotes mass
• v denotes velocity

$\orange{\bigstar}$  Solution  $\green{\bigstar}$

• m = 15 kg
• s = 160 cm = 1.6 m
• u = 0 m/s
• a = 10 m/s²

Apply 3rd equation of motion ,

$\bf \blue{\bigstar\ \; v^2-u^2=2as}\\\\\to \rm v^2-(0)^2=2(10)(1.6)\\\\\to \rm v^2=32\\\\\to \bf v=4\sqrt{2}\\\\\leadsto \bf \red{v=5.65\ m/s\ \; \bigstar}$

Apply formula for K.E ,

$\bf \pink{\bigstar\ \; K.E=\dfrac{1}{2}mv^2}\\\\\to \rm K.E=\dfrac{1}{2}(15)(4\sqrt{2})^2\\\\\to \rm K.E=\dfrac{1}{2}(15)(32)\\\\\leadsto \bf \green{K.E=240\ J\ \; \bigstar}$

i. v = 5.65 m/s

ii. K.E = 240 J

Answer 2

✔ Correct Question :-

A body of mass 15 kg falls from a height of 160 cm with an Acceleration of 10m/s².

Calculate:-

i) Velocity of the body.

ii) Kinetic Energy of the body when it just reaches the ground.

⚡Given :-

• Mass Of Body = 15kg.

• Distance = 160 cm = 1.6m.

• Initial Velocity = 0m/s.

• Acceleration = 10m/s²

⚡To Find :-

• Velocity And Kinetic Energy.

⚡Solution :-

(i) We know that,

v² - u² = 2as.

Put the values.

➨ v² - 0² = 2 × 10 × 1.6

➨ v² = 32.

➨ v = 5.65m/s²

(ii) We know that,

Kinetic Energy = ½ × m × v²

Put the values.

➨ 1/2 × 15 × (5.65)²

➨ 1/2 × 15 × 32.

➨ 15 × 16.

➨ 240 J.

Hence, These are your answers.