A 1300 kg car accelerates at 1.6 m/ s 2 [E]. A frictional force of 3800 N [W] is acting on the car. a) Draw the FBD of the car. b) Determine the applied force acting on the car.

Answer 1

a) The free-body diagram of the car can be represented as follows:

<-- Ffric -->
|
-------------
| |
| Car |
| |
-------------
|
--> Fapplied -->

Where:
- Ffric is the frictional force acting on the car (3800 N [W])
- Fapplied is the applied force on the car (unknown)

b) To determine the applied force acting on the car, we need to consider the net force acting on the car. The net force is the vector sum of all the forces acting on the car. In this case, the net force can be calculated as:

Net force = applied force - frictional force

Net force = ma (where m is the mass of the car and a is the acceleration)

Therefore, we can rewrite the equation as:

ma = Fapplied - Ffric

Solving for the applied force, we have:

Fapplied = ma + Ffric

Plugging in the given values:

Fapplied = (1300 kg)(1.6 m/s^2) + 3800 N

Fapplied = 2080 N + 3800 N

Fapplied = 5880 N

Therefore, the applied force acting on the car is 5880 N.