Maximizing Acceleration on an Icy Hill: A Work and Energy Approach

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When analyzing a car on an icy hill, the acceleration differs based on whether the driver applies the brakes or allows the car to slip freely. If the brakes are held, the car behaves as a solid body, limiting its acceleration. In contrast, if the car slips freely, the wheels may rotate, potentially increasing acceleration due to different forces at play. The discussion emphasizes the importance of understanding the work-energy principle and the role of friction in calculating linear acceleration. Ultimately, the scenario's complexity requires careful consideration of forces and energy dynamics.
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Consider a car (Mass m) above a hill (Slip angle is alpha) , which is covered with ice (Snow). if the car slips down the hill, in which circumstances its acceleration would be higher?
1- The driver push the break and hold it.
2- The driver let the car slips freely.
 
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Arsham said:
Consider a car (Mass m) above a hill (Slip angle is alpha) , which is covered with ice (Snow). if the car slips down the hill, in which circumstances its acceleration would be higher?
1- The driver push the break and hold it.
2- The driver let the car slips freely.

Hi Arsham! Welcome to PF!

Show us what you think, and why, and then we'll know how to help. :smile:
 


tiny-tim said:
Hi Arsham! Welcome to PF!

Show us what you think, and why, and then we'll know how to help. :smile:

The idea is simple. Just consider three situations:
1- (Holding break)The wheels cannot rotate, where the car and its wheels should be treated as a solid body, which is a simple case (I assume!)
2- (Slipping freely)The wheels can rotate but because of the small friction, they won't rotate. I don't know if this situation is possible or not, but if it is, I assume that it should be like No.1. Again considering the whole car as a rigid body and therefore the free diagram would be similar to no.1's
3-(Slipping freely)The wheels can rotate and they will. My question is in this part. How the free diagram should be sketched? what would be the applied forces? How the friction can be calculated? How the linear acceleration can be calculated?
 
energy

Hi Arsham! :smile:

(btw, it's "brake" when it's to do with a vehicle :wink:)
Arsham said:
… How the free diagram should be sketched? what would be the applied forces? How the friction can be calculated? How the linear acceleration can be calculated?

In a case like this, forces are confusing …

try using energy and work done instead …

if the wheels are rotating, how does that affect the energy and the speed and the work done by friction? :smile:
 


Thanks for the hint! :redface:

I would try to solve it with work & energy. Although I have to check my books to remember the procedures! :wink: Anyhow, many thanks for your comments and help. Will let you know if I succeed.
 
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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