A 1200 -kg car is pushing an out-of-gear 2100 -kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500 N . The rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 765 N , including the "friction" of turning the truck's drivetrain. What is the magnitude of the force the car applies to the truck?
To find the magnitude of the force the car applies to the truck, we can use Newton's second law: force = mass × acceleration. First, we need to find the acceleration of the car and the truck together.
The net force acting on both the car and the truck is given by the difference between the force applied by the car (4500 N) and the rolling friction of the truck (765 N):
net_force = 4500 N - 765 N = 3735 N
The total mass of the car and the truck together is:
total_mass = 1200 kg + 2100 kg = 3300 kg
Using Newton's second law, we can find the acceleration of the car and the truck together:
acceleration = net_force / total_mass = 3735 N / 3300 kg ≈ 1.1318 m/s²
Now, we can find the force the car applies to the truck by considering only the truck. We know that the truck has a rolling friction of 765 N and that the car and the truck have the same acceleration. The net force acting on the truck must be equal to the product of the truck's mass and the acceleration:
net_force_truck = truck_mass × acceleration = 2100 kg × 1.1318 m/s² ≈ 2376.78 N
The force the car applies to the truck must be greater than the rolling friction to accelerate the truck. Therefore, the force that the car applies to the truck should be:
force_car_to_truck = net_force_truck + rolling_friction = 2376.78 N + 765 N ≈ 3141.78 N
The magnitude of the force the car applies to the truck is approximately 3141.78 N.
To find the magnitude of the force the car applies to the truck, we need to first understand the situation.
In this scenario, we have a car pushing an out-of-gear truck with a dead battery. The car exerts a horizontal force of 4500 N on the ground through its drive wheels. Let's call this force F_car.
Now, when the car pushes against the ground, an equal and opposite force is exerted by the ground on the car (according to Newton's Third Law). The ground exerts a force on the car, which we can call -F_ground.
Since there is no horizontal acceleration for the car (we neglect rolling friction), the net force on the car in the horizontal direction should be zero. This means that the force exerted by the ground (-F_ground) should be equal in magnitude but opposite in direction to the force applied by the car (F_car).
Now, this force exerted by the ground (-F_ground) is transferred to the truck through their point of contact. Let's call this force F_truck.
Therefore, the magnitude of the force the car applies to the truck (F_truck) is the same as the magnitude of the force the ground exerts on the car (-F_ground), which is equal to 4500 N.
So, the magnitude of the force the car applies to the truck is 4500 N.
To find the magnitude of the force the car applies to the truck, we need to consider the forces acting on both the car and the truck.
Let's denote the force applied by the car to the truck as Fapplied.
1. Forces acting on the car:
- Horizontal force applied by the car's drive wheels: 4500 N
- Rolling friction (considered negligible): 0 N
2. Forces acting on the truck:
- Rolling friction (including friction of the truck's drivetrain): 765 N
- Force applied by the car (Fapplied)
Now, let's apply Newton's second law (Fnet = ma) to both the car and the truck:
For the car:
Fnet_car = Fapplied_car - Rolling friction_car
= 4500 N - 0 N
= 4500 N
For the truck:
Fnet_truck = Fapplied_truck - Rolling friction_truck
= - Fapplied + 765 N
As both the car and truck move in the same direction, Fapplied_car = -Fapplied_truck.
Therefore, we can equate the net forces for the car and truck:
Fnet_car = Fnet_truck
4500 N = - Fapplied + 765 N
Rearranging the equation, we get:
Fapplied = 4500 N - 765 N
= 3735 N
Thus, the magnitude of the force the car applies to the truck is 3735 N.