Child walking across frozen pond (friction involved)

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SUMMARY

The problem involves calculating the minimum time required for a child to walk 12 meters across a frozen pond with a static friction coefficient of 0.05. The net force acting against the child's movement is derived from the static friction, leading to an acceleration of -0.50 m/s². Using the kinematic equation d = V1(t) + 1/2(a)(t)², where initial velocity V1 is 0, the calculated time to reach the shore is approximately 22 seconds. This analysis confirms the relationship between friction, acceleration, and time in motion across a slippery surface.

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Homework Statement


A child stands on the surface of a frozen pond, 12m from the shore. If the coeff- of static friction between her boots and the ice is 0.05, determine the min. time req. for the child to walk to the shore without slipping.


Homework Equations


Ff = (u)(Fn)?


The Attempt at a Solution


In a FBD, I drew the Fn and Fg, and Ff acting against her movement. However, I am not sure whether she has an applied force or not; the question does say she is going to walk to the shore. I am not sure how I can manipulate the equations and end up getting time. We are only given displacement of 12m, and no other variable for a kinematic equation viewpoint.
 
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You are on the right track with Ff = u Fn.

Now considering this force, what is the kinematic equation that will relate distance, acceleration, and time? They did not tell you how much the child weighs. Maybe the mass cancels out. :smile:
 
Actually here I did it and this is what I got:
Further Work Done:
Net force in the x-component {Ff = -ma}
I split Ff into (mu)(m)(g) = -m(a) <-- canceled out the m's
(0.05)(9.81) = -a
a = -0.50m/s^2

Used this a, d, and assuming V1 is 0 to solve for t using a kinematics equation.
d = V1(t) + 1/2(a)(t)^2
t = 4 root 3, or roughly 22s.

Have I done it right? There is no answer at the back for this one.
 

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