Stopping distance and static friction

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SUMMARY

The discussion focuses on calculating the minimum stopping distance (dmin) for a large box of mass M moving at speed v0, with a small box of mass m on top. The coefficients of static friction (μs) and kinetic friction (μk) are crucial in determining the maximum deceleration that prevents slipping. The derived formula for dmin is expressed as dmin = -v0^2 / (2(μs/m)), which is contingent on ensuring that the units are consistent throughout the calculation. The participants confirm the necessity of matching units for the solution to be valid.

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  • Understanding of Newton's laws of motion
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  • Knowledge of unit consistency in physics equations
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Homework Statement


A large box of mass M is moving on a horizontal surface at speed v_0. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
Find an expression for the shortest distance dmin in which the large box can stop without the small box slipping.
Express your answer in terms of the variables v0, μs, and appropriate constants.

Homework Equations


F=μN
a=F/m

delta S = -v_i^2 / (2a_s)= -v_0^2 / 2a_s

The Attempt at a Solution


Question asks for minimum stopping distance, delta s, I am looking for the max deceleration that won't have greater force than the force due to static friction, μN.

dmin = -v_0^2 / 2(u_s/m)
Is this correct?
thanks for any help
 
Physics news on Phys.org
Do the units match?
Without matching units it cannot be right.
 

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