WEBVTT
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so here the greatest deceleration is provided by the maximum
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frictional force in this case, the normal forces equaling
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MG. And this is simply from the from the
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figure and the Newton. Using Newton's second law,
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we can say that the acceleration would be equaling the
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maximum static frictional force divided by and the Mass.
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And this would be equaling to the mat to the
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er coefficient of static friction multiplied by the normal force
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or, in this case, MG divided by EMS
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, the EMS cancel out. And this is giving
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us this coefficient of static friction multiplied by the acceleration
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due to gravity. And we can say that the
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shortest distance to stop would be that magnitude of this
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of the displacement in the ex direction. This would
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be equaling the velocity squared, divided by two times
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the acceleration. And in this case, it's me
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. The velocity squared, divided by two times the
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coefficient of static friction times G. And in this
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case we have a velocity in kilometers per hour.
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So we'll say 48 kilometers per hour, multiplied by
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1000 meters for every one kilometer multiplied by one hour
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for every 3600 seconds. We're going to square this
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term. This will be divided by two multiplied by
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the coefficient of static friction of 20.25 multiplied by 9.8
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meters per second squared. And we find that the
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magnitude of the displacement in the ex direction is going
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to be equaling two, approximately 36 meters. That
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is the end of the solution. Thank you for
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watching.