Calculating Displacement with Friction and Air Resistance

AI Thread Summary
A car rolling down a hill reaches a velocity of 54.22 m/sec before a fan applies a drag force of 250N, alongside a friction coefficient of 0.1 from the tires. The initial calculation for distance traveled without drag resulted in 27.66m, derived from kinetic energy and friction work. To accurately calculate the stopping distance, both drag force and friction must be considered together, as they both oppose the car's motion. The total force acting against the car includes both the frictional force and the drag force, necessitating a combined approach for the calculations. Understanding how to integrate these forces will lead to the correct stopping distance.
ImAlwaysConfu
Messages
3
Reaction score
0

Homework Statement



A car is rolling down a hill, it has a velocity of 54.22m/sec at the bottom of the hill. When it reaches the bottom of the hill, a fan begins to blow on it causing a drag force of 250N. Also, the tires create a friction coefficient of .1. The car eventually stops due to the Drag Force/Friction. Calculate how far the car will travel before stopping.



I can't figure out how to take into account the drag force. I know that before taking drag force into account, the car can travel 27.66m. I'm not sure if that helps or not. All I need to know is what formula to use and where I implement the drag force! Also, my homework is due TOMORROW AT 3PM EST. Any help before then will be awesome.
 
Physics news on Phys.org
How did you come to the 27.66 m? show your calcs without the drag force and I'll help with implementing the drag force...
 
First, I calculated for the Kinetic Energy which would be 1/2mv^2. 1/2(1700kg)(54.22m/sec) = 46,087J. Next, I calculated the impact friction would have: f=u(coeficcient of friction)F .1(1700)(9.8) = 1,666N. Then I calculted the work that would be done: Work = F(displacement) 46087J = 1,666N(displacement) then I divided 46087 by 1666 and came with 27.66M. If this is incorrect, could you tell me how to do it please? Thanks!
 
I think if I tell you that you would realize how easy it is. So you have an object working to the right with kinetic energy = 0.5*mv^2 your answer here is incorrect, do the calculation again... Now what is the total forces working against the vehicle? friction and drag, draw a FBD for your vehicle... Your methodology is good and accurate just combine everything now... The total force on the car is not just friction?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top