Friction Force & Direction for 925 kg Car at -12.2m/s2 Acceleration

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The discussion focuses on calculating the frictional force acting on a 925 kg car decelerating at -12.2 m/s². The normal force is determined to be 9074.25 N, and participants suggest using a free-body diagram to visualize the forces at play. The frictional force can be expressed as F_f = u * F_n, where u is the coefficient of friction. While the coefficient of friction is not directly provided, it is implied that finding it would allow for calculating the frictional force. The main emphasis is on solving for the frictional force and its direction based on the given acceleration and mass.
drinkingstraw
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The average rate of acceleration for a breaking car was -12.2m/s2. If the car had a mass of 925 kg, find the frictional force and direction.

a = -12.2m/s2
m = 925 kg
Fn = 9074.25 N

This is what I've figured out. I don't know how to find the coefficient of friction. If I get the coefficient I think I can multiply it by the normal force to get the frictional force.
 
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You're complicating the problem unnecessarily.

The question asks only for the frictional force and direction. Try drawing a free-body diagram for the car and then write down the F=ma equations.
 
F_f = u*F_n

F_n = mgF_f = u *mg

find u
 
The problem only asks for F_f :smile:
The free body diagram is the way to go. Fn and Fg cancel out. What other force or forces provide the acceleration?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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