5.1.2 Work Energy Theorem feat. Weightlifters

Question:

Comment on the work done by these four weightlifters (on the barbells).

Answer:

Remember that the work done W by a force F on an object also depends θ, which is the angle between the displacement s (of the object) and F.

W=F s\cos \theta

All four weightlifters were exerting a vertically upward force. But depending on the actual displacement of the barbell, the work done could be positive, negative or zero.

(1) \theta =0{}^\circ

The bar bell is being lifted upward by the upward force. As the formula suggests, positive work is done on the bar bell. This makes sense since the barbell gains GPE as a result.

(2) \theta =180{}^\circ

The barbell is being lowered despite the upward force. As the formula suggests, negative work is done on the barbell. This makes sense since the barbell loses GPE as a result.

(3) s=0

The bar bell reamins stationary despite the upward force (exerted by the weightlifter). Zero work is done on the bar bell. This makes sense since the barbell’s energy remains constant.

(4) \theta =90{}^\circ

The barbell is moving horizontally and importantly perpendicularly to the the upward force. As the formula suggests, zero work is done on the barbell. This makes sense because the barbell was moving horizontally at a constant speed at a constant height. The barbell maintains its KE and GPE. The kid was merely supporting the weight of the barbell. Barbell’s energy remains constant.

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