(a) How much work is done in lifting a 1.1-kg book off the floor to put it on a desk that is 0.6 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 7 ft off the ground? Solution (a) The force exerted is equal and opposite to that exerted by gravity, so the force is d²s at² mg (1.1)(9.8)= and then the work done is W=Fd= F=m (0.6)=[ (b) Here the force is given as F 21 lb, so the work done is W=Fd=21-7- ft-lb.

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
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(a) How much work is done in lifting a 1.1-kg book off the floor to put it on a desk that is 0.6 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 7 ft off the ground?
Solution
(a) The force exerted is equal and opposite to that exerted by gravity, so the force is
d²
at²
mg (1.1)(9.8) -
F=m
and then the work done is
W = Fd=
(0.6)=[
(b) Here the force is given as F 21 lb, so the work done is
WFd 21-7-
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 1.1-kg book off the floor to put it on a desk that is 0.6 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 7 ft off the ground? Solution (a) The force exerted is equal and opposite to that exerted by gravity, so the force is d² at² mg (1.1)(9.8) - F=m and then the work done is W = Fd= (0.6)=[ (b) Here the force is given as F 21 lb, so the work done is WFd 21-7- 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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