Question

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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)​

Answer 1

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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