Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops sud- denly in an accident or even in braking. Assume, for exam- ple, a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.500 with the truck bed. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution?

a) There is no relative motion between the truck and the load if they have the same acceleration.

b) mass of the load and the truck are unncessary  

Explanation: a) There is no relative motion between the truck and the load if they have the same acceleration.

Let their accelerations be a. Then a is in the direction opposite to that of the velocity of the truck (only then can the truck stop).

Let the velocity of the truck be towards right and the acceleration be towards left.

Now make the free-body diagram for the load :

Weight (mg) acts vertically downwards

Normal reaction (N) acts vertically upwards

So N = mg

Force of friction (f) acts towards left (since the block has a tendency to slip towards right)

Maximum value of friction , f = uN = umg

So f = ma

or a = ug = 4 m/s^2

Now consider the truck :

Initial velocity, u = 11 m/s

Final velocity, v = 0 m/s

a = - 4m/s^2                       (- sign shows that a is in the direction opposite to that of u)

So

s = (v^2 - u^2)/2a = 15.125 m

(b) Mass of the load and the truck are unnecessary.

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

f=ma newton

f=100x2.5=250n