Friction/Dynamic Equilibrium Help

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To determine the shortest stopping distance for a pickup truck carrying a steel file cabinet without the cabinet sliding, the coefficient of static friction for steel on steel is essential. The coefficient is identified as 0.80, allowing for the calculation of the necessary deceleration to prevent sliding. The net force acting on the cabinet is equal to the static friction force, which must counteract the cabinet's inertia. By applying the equation ma = μmg and using kinematics, the stopping distance is calculated to be 14.3 meters. This solution effectively addresses the problem using the principles of friction and motion.
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A pickup truck with a steel bed is carrying a steel file cabinet. If the truck's speed is 15 m/s, what is the shortest distance in which it can stop without the file cabinet sliding?

*---*--*-*
v(i)=15m/s
v(f)=0
Need to find x(f)

Friction: f(s) should be F(net) = 0?

Not sure where to start.
 
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Are you given a coefficient of friction for the cabinet?
Are you given any other information, such as masses?

I just reread the question:

I think you need to find the coefficient yourself, for steel on steel.
 
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Coeffiecient of friction is not given on the problem, however I did find a steel-on-steel f(s) in a table in our text --> where f(s) = 0.80
 
I am pretty sure the question should be solved as such:

The only force on the cabinet is the static friction.
Due to the cabinet's inertia (since it is already moving forward), the net force is not equal to zero. We want the friction force to move the cabinet in the opposite direction of motion so that it will not slide.

Net force on the cabinet:
Fnet = Fs

ma = μmg
a = μg

Use a kinematics equation to solve for distance, using the calculated acceleration and the known change in velocity.
 
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I came up with 14.3m - which is correct. Thanks.
 
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