Question
Asked Nov 4, 2019
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A box of mass m = 0.31 kg is set against a spring with a spring constant of k1 = 549 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 444 N/m.

a.) How far d2, in meters, will the second spring compress when the box runs into it?

b.) How fast v, in meters per second, will the box be moving when it strikes the second spring?

c.) Now assume friction is present on the surface in between the ends of the springs at their equilibrium lengths, and the coefficient of kinetic friction is μk = 0.5. If the distance between the springs is x = 1 m, how far d2, in meters, will the second spring now compress?

 

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Expert Answer

Step 1

a)

The required distance is,

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(5490.1-(4) dz 0.11 m

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Step 2

b)

The required velo...

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1 2 (549X0.1)-031) v 4.21m/s

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Science

Physics

Mechanical Properties of Matter

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