The single cable supporting an unoccupied elevator breaks when the elevator is at rest at the top of a 155 m high building. With what speed does the elevator hit the ground?
vf^2=2*g*height
so its 2*9.81*155?
K.E. = P.E.
1/2mv^2 = mgh
v^2 = 2gh
v = (2gh)^(1/2)
v = (2x9.81x155)^(1/2)
v = 55.146m/s
OH!! so that's where i went wrong. thanks again! ^w^
lol, you are welcome
but what if it asks how long it will take to fall?
To find the speed with which the elevator hits the ground, we can use the principle of conservation of energy.
First, let's find the potential energy of the elevator at the top of the building. The potential energy (PE) is given by the formula PE = mgh, where m is the mass of the elevator, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the building.
Since the elevator is at rest, it does not have any kinetic energy (KE) at the top. However, when it hits the ground, all the potential energy will be converted to kinetic energy.
At the bottom of the building, the potential energy is zero because the elevation is zero. Therefore, the kinetic energy (KE) of the elevator when it hits the ground is equal to the initial potential energy.
Using the formula for kinetic energy KE = 0.5mv^2, where v is the velocity of the elevator, we can equate the initial potential energy to the final kinetic energy:
PE = KE
mgh = 0.5mv^2
Canceling out the mass 'm' and rearranging the equation, we get:
gh = 0.5v^2
Solving for v, we have:
v^2 = 2gh
Taking the square root of both sides, we get:
v = √(2gh)
Now, we can substitute the known values into the equation. The acceleration due to gravity (g) is approximately 9.8 m/s^2, and the height of the building (h) is 155 m.
v = √(2 * 9.8 * 155)
Calculating this expression, we find:
v ≈ 17.67 m/s
Therefore, the elevator hits the ground with a speed of approximately 17.67 m/s.