a man pushes a 26kg lawnmower along the horizontal ground. He pushes down on the

Question

a man pushes a 26kg lawnmower along the horizontal ground. He pushes down on the handle at an angle of 28 degrees to the horizontal. The coefficient of static fraction of the mower with the ground is .3 and the coefficient of kinetic friction is .2 a) Draw a sketch of the situation and force diagram of the mower, labeling all forces. B) With what force does the man have to push to start the mower moving from rest? C) If he continues to push with this force after the mower is moving what will be the mowers acceleration? D) How long will he have to push with this force until it is moving at 2m/s?

Explanation / Answer

Part a)

Draw a mower with the applied force along the handle, the frictional force acting backwards from the direction of travel, a normal force (the weight of the mower) acting downward, and a force of the earth pushing up balancing the weight.

Part B)

The applied force must balance friction

Fa = Ff

Fa(cos 28) = u(mg + Fasin 28)

.883Fa = .3[(26)(9.8) + .469Fa)]

.742 Fa = 76.44

Fa = 103 N

Part C)

Net force = ma

(103)(cos 28) - .2[(26)(9.8) + (103)(sin 28)] = 26(a)

a = 1.17 m/s2

Part D)

vf = vo + at

2 = 0 + (1.17)(t)

t = 1.72 sec

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