Question

The rear side of a truck is open and a box of 40 kg mass is placed 5 m 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//s^(2) At what distance from the starting point does the box fall off the truck ? Ignore the size of the box .

Answer 1

mass of the box m =40 kg

coefficient of friction =5m

as per newton second law of motion the force on the box caused by the accelerated motion of the truck is given by

F=ma

=40×2=80n

as per newton third law of motion a reaction force of 80n is acting on the box in the backward direction the backward motion of the box and the floor of the truck this truck this force is given by

f = umg

=0.15×40×10=60n

net force acting on the block

Fnet=80-60=20/40=0.5ms 2

aback=Fnet/m =20/40=0.5ms

time is calculated as

s'=ut+(1/2)×2×(÷20)2

=20m

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