A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
1 answer:
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A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
b) The work done by the frictional force against the motion of the block is equal to:
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (
), and part of it becomes increase of thermal energy of the floor (
). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
So, we have
b) The work done by the frictional force against the motion of the block is equal to:
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (
c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
So, we have