Calculating work done in exercise movements.?

I'm trying to workout how much work is done when swinging a kettlebell from between the legs to overhead height. The movement could be traced as a semi-circle on a straight line pretty much.
Any formula I could use to calculate the work done?
I'm not too clued up on these things yet.

Am I right in saying that the kinetic energy should equal the potential energy at rest?
As far as I understand it (using a 20kg Kb as an example starting at, let's say, 30cm above the ground), potential energy would be

m.g.h = (20)(9.8)(.3)
= 58.8 Joules

Or am I barking up the wrong tree?

The force F of gravity acting on a mass b is F = bg, where g is the acceleration due to gravity, which is g = 9.8 m/s². The work W done by moving an object d meters against a force F is W=dF.
So you're right, the work done in moving the ball of mass b through d meters vertically against gravity is
W = dbg = 0.3*20*9.8 kg*m²/s²
= 58.8 joules.

By the way, 1 dietary calorie is 4185 joules, so if you lifted the kettlebell with perfect efficiency through 30cm of height, then you burned 58.8/4185 = 0.014 calories. Of course, the work wasn't done with perfect efficiency.


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One Response to “Calculating work done in exercise movements.?”

  1. Euler Says:

    The force F of gravity acting on a mass b is F = bg, where g is the acceleration due to gravity, which is g = 9.8 m/s². The work W done by moving an object d meters against a force F is W=dF.
    So you're right, the work done in moving the ball of mass b through d meters vertically against gravity is
    W = dbg = 0.3*20*9.8 kg*m²/s²
    = 58.8 joules.

    By the way, 1 dietary calorie is 4185 joules, so if you lifted the kettlebell with perfect efficiency through 30cm of height, then you burned 58.8/4185 = 0.014 calories. Of course, the work wasn't done with perfect efficiency.
    References :

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