(a) How much work is done in lifting a 1.3-kg book off the floor to put it on a desk that is 0.6 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 6 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 = (0.6) = N J. (b) Here the force is given as F = 21 lb, so the work done is W = Fd = 21.6= 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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(a) How much work is done in lifting a 1.3-kg book off the floor to put it on a desk that is 0.6 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 6 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 =
(0.6) =
N
(b) Here the force is given as F = 21 lb, so the work done is
W = Fd = 21.6=
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 0.6 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 6 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 = (0.6) = N (b) Here the force is given as F = 21 lb, so the work done is W = Fd = 21.6= 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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