Question-170120

Question Number 170120 by mr W last updated on 16/May/22

Commented by mr W last updated on 17/May/22

A truck with mass M is moving on a horizontal and straight road with speed v. A small box with mass m lies at the end of the luggage compartment of length L. The friction coefficient between tires of truck and the road is 𝛍, and that between the box and the truck is 𝛍/2. The truck obtains a full brake and it stops exactly at the instant as the box reaches the other end of the luggage compartment. Find the speed of the truck as it starts to brake.

$${A}\:{truck}\:{with}\:{mass}\:\boldsymbol{{M}}\:{is}\:{moving}\:{on} \\ $$$${a}\:{horizontal}\:{and}\:{straight}\:{road}\:{with} \\ $$$${speed}\:\boldsymbol{{v}}.\:{A}\:{small}\:{box}\:{with}\:{mass}\:\boldsymbol{{m}}\:{lies} \\ $$$${at}\:{the}\:{end}\:{of}\:{the}\:{luggage}\:{compartment} \\ $$$${of}\:{length}\:\boldsymbol{{L}}.\:{The}\:{friction}\:{coefficient} \\ $$$${between}\:{tires}\:{of}\:{truck}\:{and}\:{the}\:{road} \\ $$$${is}\:\boldsymbol{\mu},\:{and}\:{that}\:{between}\:{the}\:{box}\:{and}\:{the} \\ $$$${truck}\:{is}\:\boldsymbol{\mu}/\mathrm{2}.\:{The}\:{truck}\:{obtains}\:{a}\:{full}\: \\ $$$${brake}\:{and}\:{it}\:{stops}\:{exactly}\:{at}\:{the} \\ $$$${instant}\:{as}\:{the}\:{box}\:{reaches}\:{the}\:{other} \\ $$$${end}\:{of}\:{the}\:{luggage}\:{compartment}. \\ $$$${Find}\:{the}\:{speed}\:{of}\:{the}\:{truck}\:{as}\:{it}\:{starts}\: \\ $$$${to}\:{brake}. \\ $$

Commented by ajfour last updated on 18/May/22

$${i}\:{had}\:{again}\:{just}\:{now}\:{sir},\:{but} \\ $$$${got}\:{deleted},\:{shall}\:{answer}\:{soon} \\ $$$${again}.. \\ $$

Answered by mr W last updated on 19/May/22

N_T =(M+m)g f_T =μN_T =μ(M+m)g N_B =mg f_B =((μN_B )/2)=((μmg)/2) ma_B =f_B =((μmg)/2) ⇒a_B =((μg)/2) Ma_T =−f_B +f_T =−((μmg)/2)+μ(M+m)g ⇒a_T =μ(1+(m/(2M)))g say the truck traveled a distance d in time t, then the box traveled a distance of d+L in this time. t=(v/a_T ) d=(v^2 /(2a_T )) d+L=vt−((a_B t^2 )/2) (v^2 /(2a_T ))+L=(v^2 /a_T )−(a_B /2)((v/a_T ))^2 2La_T =(1−(a_B /a_T ))v^2 2Lμ(1+(m/(2M)))g=(1−(((μg)/2)/(μ(1+(m/(2M)))g)))v^2 μgL(2+(m/M))=(((1+(m/M))/(2+(m/M))))v^2 let ξ=(m/M) μgL(2+ξ)=(((1+ξ)/(2+ξ)))v^2 ⇒v=(2+ξ)(√((μgL)/(1+ξ)))

$${N}_{{T}} =\left({M}+{m}\right){g} \\ $$$${f}_{{T}} =\mu{N}_{{T}} =\mu\left({M}+{m}\right){g} \\ $$$${N}_{{B}} ={mg} \\ $$$${f}_{{B}} =\frac{\mu{N}_{{B}} }{\mathrm{2}}=\frac{\mu{mg}}{\mathrm{2}} \\ $$$${ma}_{{B}} ={f}_{{B}} =\frac{\mu{mg}}{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{B}} =\frac{\mu{g}}{\mathrm{2}} \\ $$$${Ma}_{{T}} =−{f}_{{B}} +{f}_{{T}} =−\frac{\mu{mg}}{\mathrm{2}}+\mu\left({M}+{m}\right){g} \\ $$$$\Rightarrow{a}_{{T}} =\mu\left(\mathrm{1}+\frac{{m}}{\mathrm{2}{M}}\right){g} \\ $$$${say}\:{the}\:{truck}\:{traveled}\:{a}\:{distance}\:{d} \\ $$$${in}\:{time}\:{t},\:{then}\:{the}\:{box}\:{traveled} \\ $$$${a}\:{distance}\:{of}\:{d}+{L}\:{in}\:{this}\:{time}. \\ $$$${t}=\frac{{v}}{{a}_{{T}} } \\ $$$${d}=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{a}_{{T}} } \\ $$$${d}+{L}={vt}−\frac{{a}_{{B}} {t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}{a}_{{T}} }+{L}=\frac{{v}^{\mathrm{2}} }{{a}_{{T}} }−\frac{{a}_{{B}} }{\mathrm{2}}\left(\frac{{v}}{{a}_{{T}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{La}_{{T}} =\left(\mathrm{1}−\frac{{a}_{{B}} }{{a}_{{T}} }\right){v}^{\mathrm{2}} \\ $$$$\mathrm{2}{L}\mu\left(\mathrm{1}+\frac{{m}}{\mathrm{2}{M}}\right){g}=\left(\mathrm{1}−\frac{\frac{\mu{g}}{\mathrm{2}}}{\mu\left(\mathrm{1}+\frac{{m}}{\mathrm{2}{M}}\right){g}}\right){v}^{\mathrm{2}} \\ $$$$\mu{gL}\left(\mathrm{2}+\frac{{m}}{{M}}\right)=\left(\frac{\mathrm{1}+\frac{{m}}{{M}}}{\mathrm{2}+\frac{{m}}{{M}}}\right){v}^{\mathrm{2}} \\ $$$${let}\:\xi=\frac{{m}}{{M}} \\ $$$$\mu{gL}\left(\mathrm{2}+\xi\right)=\left(\frac{\mathrm{1}+\xi}{\mathrm{2}+\xi}\right){v}^{\mathrm{2}} \\ $$$$\Rightarrow{v}=\left(\mathrm{2}+\xi\right)\sqrt{\frac{\mu{gL}}{\mathrm{1}+\xi}} \\ $$

Commented by ajfour last updated on 17/May/22