What is the energy of a rock with a mass of 10.2 kg on a cliff that is 300 m height?
1 answer:
The potential energy of the rock is 30,000 J
Explanation:
The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):
where
PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field
KE is the kinetic energy, which is the energy possessed by the object due to its motion
In this problem, the rock is at rest, so its kinetic energy is zero:
KE = 0
Therefore, the energy of the rock is just equal to its potential energy, which is:
where
m = 10.2 kg is the mass of the rock
is the acceleration of gravity
h = 300 m is the height of the rock above the ground
Substituting and solving, we find
Learn more about potential energy:
#LearnwithBrainly
You might be interested in
Answer:
(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact is 77.19°
Explanation:
The parameters given are;
Velocity of the plane, vₓ = 39.0 m/s
Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m
(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;
h = u·t + 1/2×g×t²
Where:
g = Acceleration due to gravity = 9.81 m/s²
u = Initial vertical velocity = 0 m/s
Hence;
1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²
∴ t = √(1500/4.905) = 17.49 s
The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m
The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The vertical velocity, , of the package just before landing is given by the relation;
² = u² + 2·g·h
u = 0 m/s
∴ ² = 0 + 2×9.81×1500 = 29430 m²/s²
= √29430 = 171.55 m/s
Hence the horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact, θ, is given as follows;
∴ θ = tan⁻¹(4.4) = 77.19°.
Answer: 247.67 V
Explanation:
Given
Potential At A
Potential at
when particle starts from A it reaches with velocity at Point while when it starts from C it reaches at point B with velocity
Suppose m is the mass of Particle
Change in Kinetic Energy of particle moving under the Potential From A to B
Change in Kinetic Energy of particle moving under the Potential From C to B
Divide 1 and 2 we get
on solving we get