You are watching: To develop muscle tone a woman lifts

This is College Physics Answers via Shaun Dychko. A womale is going to lift a weight through an angle of 60 levels, by using a force of 750 newloads at a point on this forearm that"s 2.00 centimeters away from this elbow joint so the lever arm for this muscle is 2.00 centimeters. We transform the 2.00 centimeters right into meters and we also are provided the minute of inertia of the arm—0.25 kilogram meter squared— the weight is 24.0 centimeters from the elbow and also the weight has actually a mass of 2.00 kilograms. Okay! So the first component is to figure out what the angular acceleration of this weight is going to be so we know that net torque is going to equal minute of inertia mutliplied by angular acceleration so we can deal with for α by splitting both sides by I and we gain that α is τ net over I and the torque have the right to be replaced with the pressure used by the arm multiplied by its lever before arm and also these are perpendicular so we do not have to multiply by cosine below. Then we recognize that the moment of inertia of the arm will be... well, the minute of inertia of this totality system— arm and weight included— is going to be the moment of inertia of the arm plus the moment of inertia of the weight but the weight is a single-suggest mass and also so we deserve to use this formula: mass times distance from the pivot squared. So we replace I with all of this below and then plug in numbers so the angular acceleration then is 750 newlots times 0.0200 meters divided by 0.250 kilogram meters squared— minute of inertia of the arm alone— plus—moment of inertia of the weight— 2.00 kilograms times 0.240 meters— distance from the elbow—squared and that is 41.1 radians per second squared. The net occupational done is the net torque applied multiplied by the angular displacement. Now this isn"t strictly true bereason this net torque transforms as the mass moves up but I think what we are interested in below is the work-related done by the muscle and so that"s what this is and also the muscle is constantly using 750 newloads of pressure so the torque used by the muscle doesn"t readjust. Okay! So we have the pressure applied by the muscle multiplied by its lever before arm multiplied by the angular displacement converted right into radians. So we multiply 60 levels by π radians for eincredibly 180 degrees and also the degrees cancel and also so it"s 60 times π divided by 180 and we finish up through 15.7 joules of work done by the muscle.

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