He coefficent of static friction between the floor of atruck and a box resting on it is 0.30. The truck is travelingat 80.0 km/hr. What is the least distance in which the truckcan stop and ensure that the box does not slide?

83.97 m

Explanation:

Without any other external force causing motion in the horizontal direction, it is the frictional force that brings the box to rest. The frictional force has to match the force due to deceleration experienced by the truck to not move.

So,

Frictional force = ma

Then, we calculate frictional force now,

In the vertical direction, the force balance has the weight and the normal reaction equal

N = W = mg

And frictional force = μN = μmg

where μ is the coefficient of friction = 0.3

ma = μmg

a = μg = 0.3 × 9.8 = 2.94 m/s²

Then, using the equations of motion, we obtain the distance over which this deceleration stops the truck

u = initial velocity of the truck = 80 km/h = 22.22 m/s

v = final velocity of the truck = 0 m/s, since the truck comes to rest

a = - 2.94 m/s² (negative sign because it's a deceleration)

x = distance covered during this deceleration = ?

v² = u² + 2ax

0² = 22.22² + 2(-2.94)(x)

5.88x = 493.73

x = 83.97 m

Explanation:

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