A block of mass m is dropped from the fourth floor of an office building and hits the sidewalk below at speed v. From what floor should the mass be dropped to double that impact speed?
To determine from which floor the block should be dropped to double the impact speed, we need to understand the relationship between the height of the drop and the impact speed.
Let's assume the block is dropped from a height h. We can use the concept of conservation of energy to find the impact speed.
When the block is dropped, it initially has potential energy (PE) due to its height. As it falls and reaches the ground, this potential energy is converted into kinetic energy (KE). The formula for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
At the moment of impact, all the potential energy is converted to kinetic energy. The formula for kinetic energy is KE = (1/2)mv^2, where v is the velocity.
To double the impact speed, we need to find the height from which the block should be dropped to have twice the potential energy, hence twice the impact speed.
Let's denote the new height as H. We know that the impact speed is directly proportional to the square root of the height, so doubling the impact speed is equivalent to quadrupling the potential energy.
Setting up the equation:
(mgH) = 4(mgh)
mgH = 4mgh
Dividing both sides by mg:
H = 4h
This equation tells us that the block should be dropped from a height four times greater than the original height.
Therefore, to double the impact speed, the block should be dropped from the fourth floor multiplied by four, which is the sixteenth floor.