Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.
Use this equation to determine each power value:
P = W÷Δt
P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.
A) Determining Pa:
Pa = W÷Δt
The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is
PE = mgh
PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.
We can equate W = PE = mgh, therefore we can make the following substitution:
Pa = mgh÷Δt
Given values:
m = 100.0kg
g = 9.81m/s²
h = 2.25m
Δt = 0.151s
Plug in the values and solve for Pa
Pa = 100.0×9.81×2.25÷0.151
<u>Pa = 14600W</u> (watt is the SI derived unit of power)
B) Determining Pb:
Let us use our new equation derived in part A to solve for Pb:
Pb = mgh÷Δt
Given values:
m = 150.0kg
g = 9.81m/s²
h = 1.76m
Δt = 0.052s
Plug in the values and solve for Pb
Pb = 150.0×9.81×1.76÷0.052
<u>Pb = 49800W</u>
C) Determining Pc:
Pc = mgh÷Δt
Given values:
m = 200.0kg
g = 9.81m/s²
h = 1.50m
Δt = 0.217s
Plug in the values and solve for Pc
Pc = 200.0×9.81×1.50÷0.217
<u>Pc = 13600W</u>
D) Determining Pd:
Pd = mgh÷Δt
Given values:
m = 250.0kg
g = 9.81m/s²
h = 1.25m
Δt = 0.206s
Plug in the values and solve for Pd
Pd = 250.0×9.81×1.25÷0.206
<u>Pd = 14900W</u>
Compare the following power values:
Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W
Pc is the lowest value.
Therefore, weightlifter C delivers the least power.
