ic_sol_w06d3_1 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 In-Class Problems 20-21: Work and Kinetic Energy Solutions In-Class-Problem 20 Calculating Work Integrals a) Work done by Gravity Near the Surface of the Earth: Consider an object of mass m near the surface of the earth falling directly towards the center of the earth. The gravitational force between the object and the earth is nearly constant. Suppose the object starts from an initial point y 0 and moves to a final point y f closer to the earth. How much work does the gravitational force do on the object as it falls? Solution: The displacement of the body is negative, yy f y < 0 . The gravitational 0 force is given by r r F = m g = F grav y ˆ j = m g ˆ j . grav , The work done on the body is then W gravity = F gravity y ∆=− m y . y , For a falling body, the displacement of the body is negative, f y < 0 ; therefore 0 the work done by gravity is positive, W gravity = F gravity y m > 0. y y , The gravitational force is pointing in the same direction as the displacement of the falling object so the work should be positive. When an object is rising while under the influence of a gravitational force, f y > 0 . The work done by the gravitational force for a rising body is negative, ∆≡ 0 W gravity = F gravity y m < 0, y y , since gravity is pointing in the opposite direction as the object is displaced.
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This note was uploaded on 04/22/2011 for the course PHYSICS 1441 taught by Professor Spurlock during the Spring '10 term at UT Arlington.

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ic_sol_w06d3_1 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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