A baggage handler drops your 10kg suitcase onto a conveyor belt running at2.0 m/s. The materials are such that μs=0.50 and . How far is your suitcase dragged before it is riding smoothly on the belt?

Short Answer

Expert verified

0.68 m from the position it was dropped

Step by step solution

01

Step 1. Given information

We have,

The weight of suitcase is 10kg

Conveyor belt running with 2.0m/s

The materialsμs=0.50

02

Step 2. Movement in the horizontal plane 

Since we only have movement in the horizontal plane, and knowing that the kinetic friction force is the one accelerating the suitcase, we can write:

μkmg=maa=μkg

since the suitcase starts from zero initial velocity and the movement is constantly accelerating (since the accelerating force the kinetic friction one, is constant), we have

d=at22

03

Step 3. Calculate d

Let's rewrite tknowing the first equation of motion and that the initial velocity was zero

v=v0+att=va

Now we have

d=ava22d=v22ad=v22μkg

substituting our knowing values

d=222·0.3·9.8=0.68m

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