The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from . . .

Question : The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m/s² .At what distance from the starting point does the box fall off the truck? (Ignore the size of the box) [Take g=10 m/s²]

Doubt by Yana

Solution :

m=40kg

s=5m

µ=0.15

a=2m/s²

g=10 m/s²

Maximum Force of friction on the Box

f=µR f=µmg f=0.15×40×10 f=60 N Applied Force on the Box due to acceleration of the truck

F=ma F=40×2 F=80 N Net force on the Box F'=F-f F'=80-60 F'=20 N ma'=20 N [where a' is the acceleration of the the Box on a moving truck] 40×a'=20/m a'=20/40 a'=1/2 a'=0.5 m/s²

Time in which the box will cover a distance of 5m on the truck and will fall out of the truck. S=ut+½a't² 5=0×t+½(0.5)t² 5×2=0.5t² 10/0.5=t² t²=100/5 t²=20 — (1)

Distance covered by the tuch during this time s'=ut+½at² s'=0(t)+½(2)20 s'=20 m

Hence, after covering a distance of 20 m by the truck the box will fall off the truck.