Question

A truck starts from rest and has acceleration of 4 m s-2. A box is inside truck
which is at 6 m from open end of truck. The coefficient of friction between box and
truck is 0.1. How much time box will take to slip out of truck?​

Answer 1

Answer:

hope it helps

Explanation:

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.

Answer 2

The box will slip out of the truck in 2 seconds

Explanation:

Given:

The acceleration of truck is 4 m/s²

Box is 6 m from the inside end of the truck

The coefficient of friction between the truck and box is 0.1

To find out:

The time taken by the box to slip out of the truck

Solution:

Given that

a = 4 m/s²

s = 6 m

μ = 0.1

Let the mass of the box be m

When the truck is moving, in the frame of reference of truck a pseudo force ma will act on the block in the opposite direction

This force will cause the box to move

The frictional force in the opposite direction

= 0.1mg

Therefore, net force moving the box

$F=ma-0.1mg$

$\implies F=m(4-0.1\times 10)$

$\implies F=3m$ N

Net acceleration of the box

$a'=\frac{F}{m}=\frac{3m}{m}=3$ m/s²

The box starts from rest hence intial velocity = 0

Using the second equation of motion

$s=ut+\frac{1}{2}a't^2$

$6=0\times t+\frac{1}{2}\times 3\times t^2$

$\implies t^2=4$

$\implies t=2$ s

Hope this answer is helpful.

Know More:

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

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