does anybody know how much a 7 pound light fixtures weighs falling 4 feet?

Question

A friend of mine had a light fixture weighing 7 pounds fall on his head at four feet. He received a head injury, but we are trying to determine how much the fixture weighed (with gravity and all) when it landed on his head. Thanks for all answers.

Answer 1

I can appreciate your issue with computing an equivalent weight.

Weight is really (mass * gravitational force), and what you are really asking for is the force at impact converted back to weight at a standstill. Fair enough.

Not knowing the frontal area or drag coefficient (we'll plug in 1 square foot and a CD of 1), we'll say that the velocity after falling 4 feet would be about 77 feet per second. (V = sqrt ( (2 * W) / (Cd * r * A) )

An object of 7 pounds traveling at 77 feet per second and then stopping (rather abruptly!) on your friends head produces about 875J of kinetic energy. Let's further say that your friend's head was moved about 4 inches before everything came to a halt. The impact force would be about 3750 newtons. That'd be about equilvent to trying to balance 840 pounds on his head. Hard on the neck, I'd say.

Answer 2

It still weighed seven pounds, yeah.

If you're looking for how much force it struck him with, you know distance and mass (on earth, mass and weight are interchangable), so you can calculate final speed based on acceleration due to gravity. And then we can calculate kinetic energy.


Let's do the math. We'll start by converting to metric.

4 feet = 1.22 meters
7 pounds = 3.18 kg
Acceleration = g = 9.80 m/s^2
v = initial velocity = 0 m/s
v' = final velocity
d = distance

v'^2 = v^2 + 2ad

v'^2 = 2(9.80 m/s)(1.22 m) = 239.12 m^2/s^2

v' = 15.46 m/s

Kinetic Energy is what we're looking for.

KE = (1/2)mv^2

KE = 1/2*3.18kg*239.12 m^2/s^2

= 380.2 kgm^2/s^2 = 380.2 Joules.

Which is a sizable force to strike one in the head.

Answer 3

It still weighed 7 pounds.