A 4000 kg trailer is towed up a hill with a 5% grade. If the towing truck starts from rest at the bottom of the hill and reaches a speed of 12 m/s at the top of the hill, how much is the towing force. The road from the bottom to the top of the hill is straight with a length of 200 m? Ignore friction and drag.

4840

To find the towing force, we first need to find the net force acting on the trailer as it moves up the hill. The net force is equal to the sum of the gravitational force and the force applied by the towing truck.

1. Calculate the gravitational force acting on the trailer:
The gravitational force can be calculated using the formula:
F_gravity = m * g
where m is the mass of the trailer and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 4000 kg * 9.8 m/s^2 = 39200 N

2. Find the change in potential energy:
The change in potential energy is equal to the work done by the gravitational force on the trailer. Since the trailer moves up the hill, the change in potential energy can be calculated using the formula:
∆PE = m * g * h
where h is the vertical height of the hill (given as the product of the hill's grade and the length of the hill).
∆PE = 4000 kg * 9.8 m/s^2 * (5/100) * 200 m = 39200 J

3. Calculate the work done by the towing force:
The work done by the towing force is equal to the change in kinetic energy of the trailer. The change in kinetic energy can be calculated using the formula:
∆KE = 1/2 * m * v^2
where v is the final velocity of the trailer.
∆KE = 1/2 * 4000 kg * (12 m/s)^2 = 288000 J

4. Find the net force:
The net force is the difference between the work done by the towing force and the change in potential energy:
Net force = ∆KE - ∆PE = 288000 J - 39200 J = 248800 J

5. Finally, calculate the towing force:
The towing force is equal to the net force:
Towing force = 248800 N

Therefore, the towing force required to tow the 4000 kg trailer up the 5% grade hill is 248800 Newtons.