A 747 jetliner lands and begins to slow to a stop as it moves along the runway. The mass is 3.8 105 kg, its speed is 25.9 m/s, and the net braking force is 4.27 105 N.
3.8x10^5 and 4.27x10^5
When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?
To find the deceleration of the jetliner, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
The net braking force acting on the jetliner is given as 4.27 * 10^5 N. The mass of the jetliner is 3.8 * 10^5 kg. The speed of the jetliner is 25.9 m/s.
We can start by rearranging the equation for net force and acceleration:
Net force = mass * acceleration
Rearranging the formula:
Acceleration = Net force / mass
Substituting the values:
Acceleration = (4.27 * 10^5 N) / (3.8 * 10^5 kg)
Calculating the acceleration:
Acceleration = 1.1237 m/s^2
Therefore, the deceleration of the jetliner is approximately 1.1237 m/s^2.