While exercising in a fitness center, a man lies face down on a bench and lifts

Q: While exercising in a fitness center, a man lies face down on a bench and lifts

Here , a) let the angular acceleration is a Using second law of motion I * a = torque (0.9 + 10.5 * 0.28^2 ) * a = 1410 * 0.035 a = 28.64 rad/s the angular acceleration is 28.64 rad/s b) work done by muscle = I * a * theta work done by muscle = (0.9 + 10.5 * 0.28^2 ) * 28.64 * (15 * 3.141/180) work done by muscle = 12.92 J the work done by muscle ia 12.92 J

ID: 1367306 • Letter: W

Question

While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contracting the muscles in the back of the upper leg.

a) Find the magnitude of the angular acceleration produced given the mass lifted is 10.5 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is 0.900 kg·m2, the muscle force is 1410 N, and its effective perpendicular lever arm is 3.50 cm.

b) How much work is done if the leg rotates through an angle of 15.0° with a constant force exerted by the muscle?

Explanation / Answer

Here ,

let the angular acceleration is a

Using second law of motion

I * a = torque

(0.9 + 10.5 * 0.28^2 ) * a = 1410 * 0.035

a = 28.64 rad/s

the angular acceleration is 28.64 rad/s

work done by muscle = I * a * theta

work done by muscle = (0.9 + 10.5 * 0.28^2 ) * 28.64 * (15 * 3.141/180)

work done by muscle = 12.92 J

the work done by muscle ia 12.92 J

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While exercising in a fitness center, a man lies face down on a bench and lifts

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While exercising in a fitness center, a man lies face down on a bench and lifts

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