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A person pushes a 11.5KG lawn mower with an applied force of 70.0N along the 45.0-degree-handle. if the lawn mower moves at a constant velocity
Determine the net force on the mower:
Fnet=__N
Determine the horizontal component of applied force on the mower:
Fx=__N (round to 3 digits)
Determine the horizontal frictional force acting on the mower
Ff=__N (Round to 3 digits)
Calculate the applied force the person must exert on the mower for it to accelerate at a rate of 1.00m/s^2
F=__N (round to 3 digits)
Answers
Force is a vector quantity hence it can be resolved into a vertical and a horizontal components.
1) The horizontal component of the applied force is given by;
Fnet = F cos θ
Fnet = 70.0N × cos (45.0)
Fnet = 49.5 N
2) The horizontal component of the frictional force is;
F = -Ff + mgsinθ
F = applied force
Ff =mgsinθ - F
Ff = (11 × 10 × sin 45) - 49.5
Ff = 28.3 N
Horizontal component of Ff
28.3 N cos 45 = 19.9 N
3) The applied force is given by;
F = 11 Kg × 1.00m/s^2 = 11.0 N
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