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Question

A conveyor belt is moving at a constant speed of 2m/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 the belt before coming to rest on it (taking g=10m/s2), is


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Solution

Step 1: Given

Speed of conveyor belt (u)=2m/s

Coefficient of friction (μ)=0.5

Acceleration due to gravity (g)=10m/s2

Final speed =v

Step 2: Formulae used

The third equation of motion v2=u2+2aS, where v is the final speed, u is the initial speed, a is the acceleration, and S is the distance covered.

Step 3: Calculation

We know, that frictional force is the cause of deceleration of the box, thus,

F=μmgbutF=ma-ma=μmga=-μg

where μ is the coefficient of friction.

Here the negative sign indicates the deceleration of the box.

Acceleration of the box considering friction,a=-μg=-0.5×10=-5m/s2

Applying the third equation of motion

v2=u2+2aS

0=22-2×5×S0=4-10SS=410=25=0.4m

Thus, the distance the box will move with respect to the conveyor belt is 0.4 meters.


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