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(a) How much work is done in lifting a 1.3-kg book off the floor to put it on a desk that is 0.9 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s². (b) How much work is done in lifting a 21-lb weight 4 ft off the ground? olution (a) The force exerted is equal and opposite to that exerted by gravity, so the force is F = m = mg = (1.3) (9.8) = d²s dt² and then the work done is W = Fd= (0.9) = N J. (b) Here the force is given as F= 21 lb, so the work done is W = Fd = 21.4 = ft-lb. Notice that in part (b), unlike part (a), we did not have to multiply by g because we were given the weight (which is a force) and not the mass of the object.

(a) How much work is done in lifting a 1.3-kg book off the floor to put it on a desk that is 0.9 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s². (b) How much work is done in lifting a 21-lb weight 4 ft off the ground? olution (a) The force exerted is equal and opposite to that exerted by gravity, so the force is F = m = mg = (1.3) (9.8) = d²s dt² and then the work done is W = Fd= (0.9) = N J. (b) Here the force is given as F= 21 lb, so the work done is W = Fd = 21.4 = ft-lb. Notice that in part (b), unlike part (a), we did not have to multiply by g because we were given the weight (which is a force) and not the mass of the object.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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(a) How much work is done in lifting a 1.3-kg book off the floor to put it on a desk that is 0.9 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s². (b) How much work is done in lifting a 21-lb weight 4 ft off the ground? Solution (a) The force exerted is equal and opposite to that exerted by gravity, so the force is F = m = mg = (1.3) (9.8) = d²s dt² and then the work done is W = Fd = (0.9) = N J. (b) Here the force is given as F = 21 lb, so the work done is W = Fd = 21. 4 = ft-lb. Notice that in part (b), unlike part (a), we did not have to multiply by g because we were given the weight (which is a force) and not the mass of the object.
Transcribed Image Text:(a) How much work is done in lifting a 1.3-kg book off the floor to put it on a desk that is 0.9 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s². (b) How much work is done in lifting a 21-lb weight 4 ft off the ground? Solution (a) The force exerted is equal and opposite to that exerted by gravity, so the force is F = m = mg = (1.3) (9.8) = d²s dt² and then the work done is W = Fd = (0.9) = N J. (b) Here the force is given as F = 21 lb, so the work done is W = Fd = 21. 4 = ft-lb. Notice that in part (b), unlike part (a), we did not have to multiply by g because we were given the weight (which is a force) and not the mass of the object.
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Step 1: Introduce the formula of workdone

The work done is the product of force and displacement along the force direction. 


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