1. Understanding the Forces
* Gravity: The car's weight (19600 N) acts vertically downwards. We need to find the component of this force that acts parallel to the slope, which will be the force pulling the car down the incline.
* Friction: The frictional force opposes the car's motion, acting uphill.
* Net Force: The net force acting on the car is the difference between the component of gravity pulling it down and the frictional force pushing it uphill.
2. Calculating the Component of Gravity
* Since the slope is at 45 degrees, the component of gravity parallel to the slope is:
* Weight * sin(45°) = 19600 N * sin(45°) ≈ 13859 N
3. Calculating the Net Force
* Net force = Force of gravity down the slope - Frictional force
* Net force = 13859 N - 2000 N = 11859 N
4. Calculating the Acceleration
* We know the car's weight is 19600 N, so we can find its mass:
* Mass = Weight / Acceleration due to gravity (g) = 19600 N / 9.8 m/s² ≈ 2000 kg
* Now we can calculate the acceleration using Newton's second law (F = ma):
* Acceleration (a) = Net force / Mass = 11859 N / 2000 kg ≈ 5.93 m/s² (This is the deceleration since it's acting against the car's motion)
5. Converting Speed to m/s
* The car's initial speed is 63 km/h, which we need to convert to meters per second:
* 63 km/h * (1000 m / 1 km) * (1 h / 3600 s) ≈ 17.5 m/s
6. Calculating the Stopping Distance
* We'll use the following kinematic equation:
* v² = u² + 2as
* Where:
* v = final velocity (0 m/s since the car comes to rest)
* u = initial velocity (17.5 m/s)
* a = acceleration (deceleration, -5.93 m/s²)
* s = stopping distance (what we want to find)
* Rearranging the equation to solve for s:
* s = (v² - u²) / (2a)
* s = (0² - 17.5²) / (2 * -5.93) ≈ 25.9 meters
Therefore, the car will travel approximately 25.9 meters before coming to rest.
