A girl coats down a hill on a sled

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The discussion revolves around calculating the distance a sled travels on level ground before coming to a stop, given its initial speed of 7.0 m/s, a coefficient of kinetic friction of 0.050, and a total weight of 645 N. Participants suggest starting with a free body diagram to identify forces acting on the sled, including friction and weight. Key equations mentioned include those for final velocity, acceleration, and frictional force. The conversation emphasizes breaking down the problem into components to simplify calculations. Ultimately, the goal is to determine how far the sled will slide before coming to rest.
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reaching level ground at the bottom with a speed of 7.0 m/s. the coefficient of kinetic friction between the sled's runners and the hard, icy snow is 0.050, and the girl and sleed together weight 645 N. how far does the sled travel on the ground level before coming to rest?
 
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chenny said:
reaching level ground at the bottom with a speed of 7.0 m/s. the coefficient of kinetic friction between the sled's runners and the hard, icy snow is 0.050, and the girl and sleed together weight 645 N. how far does the sled travel on the ground level before coming to rest?

What have you tried so far? Where are you running into difficulty?
 
i just took one look at the problem and my mind went blank. i totally do not know how to start it.
 
chenny said:
i just took one look at the problem and my mind went blank. i totally do not know how to start it.

Draw a free body diagram. What are the forces acting on the sled?

Remember to break everything up into x- and y-components.

v_{f}^{2}=v_{i}^{2}+2ad

v_{f}=v_{i}+at

x_{f}=x_{i}+v_{0}t+\frac{1}{2}at^{2}

F_{k}=\mu_{k}N

F=ma

w=mg
 
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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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