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Standard IX

Physics

Second Law of Motion

A suitcase is...

Question

A suitcase is gently dropped on a conveyor belt moving at $3 m / s$ . If the coefficient of friction between the belt and the suitcase is $0.5$ . Find the displacement of suitcase relative to converyor belt before the slipping between the two is stopped ( $g = 10 m / s^{2}$ )

2.7 m

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1.8 m

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0.9 m

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1.2 m

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Solution

The correct option is C $0.9 m$
SInce the suitcase is slipping, retardation due to friction is given by:
$a = - \frac{μ M g}{M} = - μ g$
So
$v^{2} - u^{2} = 2 a s$
In the frame of the conveyer $v = 0$ and $u = 3 m / s$
$⟹ - 3^{2} = - 2 μ g s$
$⟹ s = 0.9 m$
Hence Option C is correct

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Similar questions

Q. A conveyor belt is moving at a constant speed of

2

m / s .

A box is gently dropped on it. The coefficient of friction between them is

μ = 0.5 .

The distance that the box will move relative to belt before coming to rest on it. taking

g = 10 {m s}^{- 2}

:

Q. A conveyor belt is moving at a constant speed of

2 m / s

, a box is gently dropped on it. The coefficient of friction between them is

0.5

. The distance that the box will move relative to belt before coming to rest on it (taking

g = 10 m s^{- 2}

), is:

A girl is standing on ground near a horizontally moving conveyor belt. She throws her suitcase horizontally. Such that the initial velocity of suitcase relative to stationary girl is along dotted line shown in top view of conveyor belt.

Which of the following arrows represents the direction of friction on the bag at the initial moment.

Which of the following arrows represents direction of friction at initial moment.

An object is gently placed on a long conveyer belt moving with a uniform speed of $5 m s^{- 1}$ . If the coefficient of friction between the block and the belt is 0.5, how far will the box move on the belt (relative to belt) before coming to rest? Take $g = 10 m s^{- 2}$ .

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A suitcase is gently dropped on a conveyor belt moving at 3m/s. If the coefficient of friction between the belt and the suitcase is 0.5. Find the displacement of suitcase relative to converyor belt before the slipping between the two is stopped (g=10m/s2)

The correct option is C 0.9mSInce the suitcase is slipping, retardation due to friction is given by:a=−μMgM=−μgSov2−u2=2as In the frame of the conveyer v=0 and u=3m/s⟹−32=−2μgs⟹s=0.9mHence Option C is correct

A suitcase is gently dropped on a conveyor belt moving at $3 m / s$ . If the coefficient of friction between the belt and the suitcase is $0.5$ . Find the displacement of suitcase relative to converyor belt before the slipping between the two is stopped ( $g = 10 m / s^{2}$ )

The correct option is C $0.9 m$
SInce the suitcase is slipping, retardation due to friction is given by:
$a = - \frac{μ M g}{M} = - μ g$
So
$v^{2} - u^{2} = 2 a s$
In the frame of the conveyer $v = 0$ and $u = 3 m / s$
$⟹ - 3^{2} = - 2 μ g s$
$⟹ s = 0.9 m$
Hence Option C is correct