Question

A crate rest on a flatbed truck which is initially traveling at 13.6 m/s on a level road. the driver applies the brakes and the truck is brought to a halt in a distance of 41 m. if the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding

Answer 1

`\begin{gathered} \uparrow+\Sigma Fy=0 \\ N-mg=0 \\ N=mg \\ \rightarrow+\Sigma Fx=ma \\ Ff=-ma \\ \text{But} \\ Ff=N\mu \\ mg\mu=-ma\text{ (1)} \\ To\text{ find a} \\ v1=13.6\text{ m/s} \\ v2=\text{ 0 m/s} \\ x=41m \\ a=\text{?} \\ v2^2=v1^2+2ax \\ \text{Solving a} \\ v2^2-v1^2=2ax \\ a=(v2^2-v1^2)/(2x) \\ a=\frac{(\text{ 0 m/s})^2-(13.6\text{ m/s})^2}{2(41m)} \\ a=-2.26m/s^2 \\ From\text{ (1) solving }\mu \\ mg\mu=-ma \\ \mu=(-ma)/(mg) \\ \mu=(-a)/(g) \\ \mu=(-(-2.26m/s^2))/(9,81m/s^2) \\ \mu=0.23 \\ \text{The value of the coefficient of friction is }0.23 \end{gathered}`