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.

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.