Question

A box sits on the back of a flatbed truck. If the coefficient of static friction between the box and the truck bed is μs = 0.400, what is the shortest distance in which the truck can stop on level ground without the box sliding if it is initially traveling at v = 15.0 m/s?

Answer 1

Answer:

28,699m

Explanation:

The force to make the box move should be μs.N=μs.m.g=m.|a|

then,

|a|=μs.g

Being

μs coefficient of static friction,

N the force made by the truck on the box caused by the gravity force,

m the mass,

g the acceleration of gravity

and a the acceleration of the truck.

$x = v \times t + \frac{1}{2} \times a \times {t}^{2}$

as the truck is stopping, the acceleration is negative. then,

$x = v \times t - \frac{1}{2} \times |a| \times {t}^{2}$

$|a| = v \div t \\ t = v \div |a|$

$x = v \times (v \div |a| ) - \frac{1}{2} \times |a| \times {(v \div |a|)}^{2}$

$x = v \times (v \div μs.g ) - \frac{1}{2} \times |a| \times {(v \div μs.g)}^{2}$

$x = \frac{{v}^{2}}{μs \times g} - \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = 0.5 \times \frac{{(15m/s) }^{2} }{0.4 \times 9.8m/ {s}^{2} } = 28.699m$

28,699m