A weight lifter bench‑presses a 130.0kg barbell from his chest to a height equal the length of his arms, 0.850 m. How much work does he do?

Question

What is the potential energy of the barbell relative to the weight lifter's chest after he lifts the barbell?

How many food calories does the weight lifter use if he does this lift 35 times? Assume the weight lifter does no work in letting the barbell drop back to his chest.

I appreciate you taking the time to help me understand this!

Answer 1

Hi Eunjin, this is question is related to potential energy.

Potential energy is the energy that arises when moving away from some sort of equilibrium, such as gravity

The equation is,

∆PE = mg∆h

In other words,

the change in potential energy

is equal to

mass times

gravitational acceleration times

the change in height of the object.

If we pay attention to the units, we have

kilograms × meters/second² × meters = kg × (meters²/seconds²) = Joules (J)

These are the units of energy! (The way I remember it is from Einstein's famous equation, E = mc²)

Work, on the other hand, is force times distance moved. It's a lot of work to generate a force over a long distance!

It is also in units of energy

W = F × d

Remember, F = mass × acceleration = kg × meters/second²

Thus, work = kg × meters/second² × meters = kg × (meters²/seconds²)

Going back to your question,

m = 130 kg

∆h = 0.85 m eters

g = 9.81 meters/second²

Looking back at our force equation (F = ma), the minimum force we need to apply in order to lift the object is the weight of the object (mass times gravitational acceleration) = 130 * 9.81 = 1,275 Newtons

Thus the work is thus the Force times distance

1,275 * 0.85 = 1,084 Joules

Interestingly, the change in potential energy is also equal to this number!

∆PE = 130 * 0.85 * 9.81 = 1084 Joules after the bar has been lifted

So you do the amount of work equal to the potential energy required to lift the object