How Much Force Does a Track Exert on a Speeding Car?

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To determine the average net force exerted by the track on a speeding car, one can use the formula F=ma, where 'm' is the mass of the car and 'a' is the acceleration. In this case, with a mass of 1550 kg and an acceleration from 10 m/s to 30 m/s over 10 seconds, the average net force can be calculated as 3100 N. The discussion clarifies that the track does indeed apply this force, as it is responsible for providing the necessary horizontal force for acceleration. The conversation also humorously touches on the idea of a frictionless track and its implications for acceleration. Overall, understanding the role of the track in exerting force is essential for solving the problem.
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Homework Statement


After a day of testing race cars, you decide to take your own 1550 kg car onto the test track. While moving down the track at 10 m/s, you uniformly accelerate to 30 m/s in 10 s. What is the average net force that the track has applied to the car during the 10 s interval?


Homework Equations


F=ma


The Attempt at a Solution


Is it just (20/10)*1550? the only thing that confuses me is "the average net force that the track has applied to the car." The track applied to the car. Did the track apply the force?
 
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Your calculation is correct. Yes, the track applied the force. What else is there (not part of the train) that could provide a horizontal force? Or, if the track were completely frictionless, i.e. cannot provide a horizontal force, would the train be able to accelerate?
 
If the dragster moves like a squid :) Sorry, I couldn't resist, but seriously, thanks!
 
Sorry, dragster not train. Rocket-propelled dragsters do act like squids, so your comment is not as silly as you might think.
 
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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