Answer:
|F| = 393750 N
Explanation:
Given that,
Total mass of the train, m = 750000 kg
Initial speed, u = 84 m/s
Final speed, v = 42 m/s
Time, t = 80 s
We need to find the net force acting on the train. The formula for force is given by :
F = ma
So, the magnitude of net force is 393750 N.
Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
Where:
d: is the distance traveled = 46.1 m
: is the final speed of the truck = 0 (it stops)
: is the initial speed of the truck = 17.9 m/s
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!