Question

An object of mass 30 kg is dropped from a height of 8 m. The potential energy and kinetic energy of the object at a height of 3m from the ground is given by __ and __(g=10m/s²) 1 point

Answer 1

Given :-

• Mass of object (m) = 30 kg
• Height (H) = 8 m (initially )
• Height (h) = 3 m ( finally )
• Acceleration due to gravity (g) = 10 m/s ²

To find :-

• PE and KE  of the object at a height of 3 m

Solution :-

Potential energy of the object at a height of 3 m ,

➜ PE ' = m g h

➜ PE ' = 30 x 10 x 3

➜ PE ' = 900 J

By applying law of conservation of momentum ,

➜ KE + PE = KE' + PE '

here ,

• KE = initial kinetic energy
• PE = initial potential energy
• KE' = final kinetic energy
• PE' = final potential energy

➜ mu² / 2 + mgH = KE'  + mgh

➜ mgH = KE'  + mgh [ ∵ The initial velocity (u) of the ball is 0 ]

➜ KE' = mgH - mgh

➜ KE' = mg [ H-h]

➜ KE' = 30 x 10 [ 8 - 3 ]

➜ KE ' = 30 x 10 x 5

➜ KE' = 1,500 J

The potential energy and kinetic energy of the object at a height of 3m from the ground is given by 900 J and 1,500 J .

Answer 2

Answer:

Explanation:

Given :-

Mass of object (m) = 30 kg

Height (H) = 8 m (initially )

Height (h) = 3 m ( finally )

Acceleration due to gravity (g) = 10 m/s ²

To find :-

PE and KE  of the object at a height of 3 m

Solution :-

Potential energy of the object at a height of 3 m ,

➜ PE = m g h

➜ PE = 30 x 10 x 3

➜ PE = 900 J
By applying law of conservation of momentum ,

➜ KE + PE = KE + PE
KE = initial kinetic energy

PE = initial potential energy

KE = final kinetic energy

PE' = final potential energy

➜ mu² / 2 + mgH = KE'  + mgh

➜ mgH = KE'  + mgh [ ∵ The initial velocity (u) of the ball is 0 ]

➜ KE = mgH - mgh

➜ KE = mg [ H-h]

➜ KE = 30 x 10 [ 8 - 3 ]

➜ KE = 30 x 10 x 5

➜ KE = 1500 J

The potential energy and kinetic energy of the object at a height of 3m from the ground is given by 900 J and 1,500 J.