Question
A 20.0 kg block is being pushed forward on a flat surface with a force of magnitude 45.0 N against a frictional force of 13.0 N
To draw this free-body diagram, would I only have the forces mentioned in the question? or would i include the -9.8m/s^2 force of gravity? Also, how would I find the change in kinetic energy if the block is pushed 4 m?
To draw this free-body diagram, would I only have the forces mentioned in the question? or would i include the -9.8m/s^2 force of gravity? Also, how would I find the change in kinetic energy if the block is pushed 4 m?
Answers
You only need the horizontal forces in this case because the friction force(not the coefficient) is given.
45 - 13 = 20 a
work done = change in kinetic energy
= (45 - 13) (4) Newton meters or Joules
45 - 13 = 20 a
work done = change in kinetic energy
= (45 - 13) (4) Newton meters or Joules
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