A tractor drags a 313kg plow across a field at 2.0 m/s . If the coefficient of friction between the plow and the ground is 0.570, how much force does the tractor apply to the plow ?

The tractor applies 1,748.418 N to the plow.

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn=m.a

If the acceleration is zero, then the net force is also zero. That means all forces acting on the mass are balanced.

The tractor drags the m=313 Kg plow at a constant speed, thus the acceleration is zero and so is the net force.

The horizontal forces acting on the plow are:

Since both forces are balanced, then:

Ft=Fr

The friction force is calculated as:

Where is the coefficient of friction between the plow and the ground and N is the normal force. In the situation described, the normal force is equal to the weight of the plow:

The normal force is N=3,067.4 N

The friction force is now calculated:

Thus, the tractor applies 1,748.418 N to the plow.

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explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

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