Question

A tractor drags a 251kg plow across a field at a constant 3.0m/s. If the coefficient of

friction between the plow and the ground is 0.505, how much force does the tractor
apply to the plow?

Answer 1

The tractor applies a force of 1241.245 newtons to drag the plow across the field at a constant speed, which is equal to the frictional force that is calculated using the coefficient of friction and the normal force.

Step-by-step explanation:

To determine the force the tractor applies to the plow, we need to consider the frictional force acting on the plow.

The force of friction (f) can be calculated using the formula: f = μ × N, where μ is the coefficient of friction, and N is the normal force. Since the plow is moving at a constant speed, the net force is zero, and hence the force applied by the tractor is equal to the frictional force.

Let's calculate the frictional force: f = μ × (m × g) where m is the mass of the plow (251 kg) and g is the acceleration due to gravity (9.8 m/s²).

Frictional force = 0.505 × (251 kg × 9.8 m/s²)

Frictional force = 0.505 × 2459.8 N

Frictional force = 1241.245 N

Therefore, the tractor must apply a force of 1241.245 newtons to drag the plow across the field at a constant speed.