Physics, asked by xghjk706, 9 months ago

The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. 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). Fig. 5.22

Answers

Answered by Anonymous
1

Given: the rear side of a truck is open and a box of 40kg mass is placed 5m away from the open end as shown in the figure. 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 2m/s

2

To find the distance (in m) traveled by truck by the time box falls from the truck. (Ignore the size of the box)

Solution:

Mass of the box, m=40kg

μ=0.15

u=

a=2m/s

2

S=5m

F=ma

⟹F=40×2=80N

This is due to acceleration of truck.

f=μmg

⟹f=0.15×40×10=60N

Therefore, net force acting on the block

F

net

=80−60=20N backward

Backward acceleration produce is

a

back

=

m

F

net

=

40

20

=0.5m/s

2

Using the second equation of motion, time t can be calculated as:

S

=ut+

2

1

a

back

t

2

⟹S=0+

2

1

×0.5×t

2

∴t=

20

S

Distance S, travelled by the truck in

20

S

S=ut+

2

1

at

2

⟹S=0+

2

1

×2×(

20

)

2

⟹S=20m

is the distance (in m) traveled by truck by the time box falls from the truck

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