A block of mass m is dropped from the fourth floor of an office building, subsequently hitting the sidewalk at speed v. From what floor should the mass be dropped to double that impact speed?
The 16th floor
To solve this problem, we can utilize the concept of conservation of mechanical energy. The potential energy of the block when it is dropped from a certain height will be converted into kinetic energy as it reaches the ground. By equating the initial potential energy with the final kinetic energy, we can find the answer.
The potential energy (PE) of an object is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and h is the height.
The kinetic energy (KE) of an object is given by the formula KE = 0.5mv², where m is the mass and v is the velocity or speed.
Initially, when the block is dropped from the fourth floor, the potential energy is equal to the initial kinetic energy:
PE_initial = KE_initial
mgh = 0.5mv_initial²
We can cancel out the mass (m) on both sides:
gh = 0.5v_initial²
Now, let's consider the situation when the block is dropped from the floor that will double the impact speed. Let's call this floor "n" (where the block is dropped from).
The height from the "n"th floor can be calculated by multiplying the height from the fourth floor (h) by (4-n). This is because each floor is equal in height, so when we drop from the "n"th floor, we are essentially dropping from a height that is (4-n) times the height from the fourth floor.
Thus, the height (h_n) from the "n"th floor is given by h_n = h * (4-n).
The potential energy (PE_n) when dropped from the "n"th floor is mgh_n = mgh * (4-n).
The final velocity (v_n) when dropped from the "n"th floor can be calculated using the formula:
0.5mv_n² = mgh_n
We can cancel out the mass (m) on both sides:
0.5v_n² = gh_n
Substituting h_n = h * (4-n):
0.5v_n² = g * h * (4-n)
Now, we can solve for "n" by equating the final kinetic energy (0.5v_n²) with double the initial kinetic energy (2 * 0.5v_initial²):
0.5v_n² = 2 * 0.5v_initial²
Simplifying:
g * h * (4-n) = 2 * g * h
Dividing both sides by g * h:
4 - n = 2
Rearranging and solving for "n":
n = 4 - 2
n = 2
Therefore, to double the impact speed, the block should be dropped from the second floor of the office building.