Dynamics please show all work A pickup truck is traveling f

Dynamics *** please show all work

A pickup truck is traveling forward at 25 m/s. The bed is loaded with boxes whose coefficient of friction with the bed is 0.40 . What is most nearly the shortest time that the truck can be brought to a stop such that the boxes do not shift ?

Solution

The Free Body Diagram of the truck loaded with boxes is given as:

By Newton’s Second Law of Motion,

Fx = ma_x

Fy = ma_y

In x-direction,

ma_x = f………………………… (1)

In y-direction,

N = mg………………………… (2)

Now, for the boxes to remain in equilibrium, the resultant force in the x-direction should be zero and this horizontal force ‘f’ is balanced by the frictional force acting on the tires of the truck in the direction opposite to that of f.

Therefore, f = N ………………………… (3)

Equating (1) and (3), we get,

ma_x = N = (mg) … [from (2)]

a_x = g = o.4*9.81 = -3.924 m/s² (negative sign for deceleration)

Now,

v = u = at

where v = final velocity of the truck = 0 (at rest)

and u = initial velocity of the truck = 25 m/s

Therefore,

0 = 25 – 3.924*t

t = 6.371 seconds