Question

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[tex]The rear side of a truck is open and a box of mass 20 Kg is placed on a truck 4 metre away from the open end.
___________❤
ü= 0.15
g=10m/s²
___________❤
The truck starts from rest with an acceleration of 2m/s² on a straight road.
At what distance the box will fall off the truck from the starting point ?

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Answer 1

||✪✪ QUESTION ✪✪||

The rear side of a truck is open and a box of mass 20 Kg is placed on a truck 4 metre away from the open end.

ü= 0.15

g=10m/s²

The truck starts from rest with an acceleration of 2m/s² on a straight road. At what distance the box will fall off the truck from the starting point ?

|| ✰✰ ANSWER ✰✰ ||

Given That :-

→ m = 40 kg

→ a = 2 m/s²

→ μ = 0.15

→ d = 4m.

Here, net force acting on the box is given by,

Net Force = Force due to speed of truck - Frictional force

→ Fnet = Fx - Ff

→ Fnet = ma - μmg

→ Fnet = 40×2 - 0.15×40×10

→ Fnet = 80 - 60

→ Fnet = 20 N

Net acceleration = Fnet/m

→ a' = 20/40 = 0.5 m/s²

So ,

Time required for box to fall off :-

s = ut + 1/2 a't²

→ 4 = 0×t + (1/2)×0.5×t²

→ 4 = 0.25t²

→ 4 = (1/4)t²

→ t² = 4 * 4

→ t = 4 seconds.

So, After 4 seconds The box will Fall of The Truck and ,,

Distance travelled by truck in 4 s :-

→ s' = ut + 1/2 at²

→ s' = 0×4 + (1/2)×2×4²

→ s' = 0 + 4²

→ s' = 16m .

Hence, The Truck will travel 16m before box falling off.

Answer 2

Explanation:

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