A pole-vaulter just clears the bar at 5.34 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3.8E3 J. What is his weight?
To determine the weight of the pole-vaulter, we need to first understand the concept of potential energy and its relationship to weight.
Potential energy is the energy possessed by an object due to its position in a gravitational field. The formula for potential energy (PE) is given by:
PE = mgh
Where:
m is the mass of the object
g is the acceleration due to gravity
h is the height or vertical displacement of the object
In this case, the change in potential energy (ΔPE) is given as -3.8E3 J (negative because the potential energy decreases as the vaulter falls).
ΔPE = PE_final - PE_initial
Since the vaulter cleared the bar at a height of 5.34 m, we can set the initial potential energy (PE_initial) to zero (as we are measuring it relative to the starting position). Therefore:
ΔPE = PE_final - 0
Simplifying:
ΔPE = PE_final
Now, we can substitute the given values into the equation:
-3.8E3 J = mgh
Rearranging the equation to solve for the weight (mg):
mg = -3.8E3 J / h
Plugging in the values:
Assuming standard gravity (g ≈ 9.8 m/s^2):
mg = -3.8E3 J / 5.34 m
mg = -712.7 kg*m^2/s^2 /m
Simplifying:
mg = -712.7 kg * m/s^2
The weight (mg) is expressed in newtons, so:
Weight = 712.7 N (approximately)
Therefore, the weight of the pole-vaulter is approximately 712.7 N.