a rock with a mass of 2 kg is thrown from the top of a cliff 75 m high with a velocity of 10m/s. how much Kinetic energy does it have when it gets to the bottom?
V^2 = Vo^2 + 2g*h
V^2 = !0^2 + 19.6*75 = 1570
V = 39.62 m/s
KE = 0.5m*V^2 = 1*39.62^2 = 1570 J.
Anser is listen
To determine the kinetic energy of the rock when it reaches the bottom of the cliff, we need to consider the conservation of energy.
First, let's calculate the potential energy of the rock when it's at the top of the cliff. The potential energy is given by the equation:
Potential Energy = mass * gravitational acceleration * height
Given:
- Mass (m) = 2 kg
- Gravitational acceleration (g) ≈ 9.8 m/s^2
- Height (h) = 75 m
Potential Energy = 2 kg * 9.8 m/s^2 * 75 m
Potential Energy = 1470 Joules
Since energy is conserved, the potential energy at the top will be converted into kinetic energy at the bottom of the cliff.
The kinetic energy is given by the equation:
Kinetic Energy = 0.5 * mass * velocity^2
Given:
- Mass (m) = 2 kg
- Velocity (v) = 10 m/s
Kinetic Energy = 0.5 * 2 kg * (10 m/s)^2
Kinetic Energy = 100 Joules
Therefore, the rock will have 100 Joules of kinetic energy when it reaches the bottom of the cliff.