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.

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day