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
Answers
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:
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.