Solving a Car Coasting Downhill Problem: Average Retarding Force

AI Thread Summary
To solve the problem of a 1200 kg car coasting downhill, the average retarding force due to friction can be determined by analyzing the energy balance and forces acting on the car. The car reaches a speed of 20 m/s after descending 35 m over 800 m. The kinetic energy at the bottom can be calculated, and the difference between this and the kinetic energy that would exist without friction indicates the work done against friction. The forces acting on the car include gravity and friction, with the normal force being irrelevant for the calculation of retarding force. Understanding the relationship between acceleration, velocity, and distance is crucial for solving the problem effectively.
TeeNaa
Messages
7
Reaction score
0
A 1200 kg car start from rest and coast down a uniform grade, at the bottom of which ti has a speed of 20 m/s. If the car has traveled 800 m along the grade and has descended 35 m, what is the average retarding force(friction) encountered by the car? How would I solve this problem? I know Fx = ma = Ff - mgsintheta Fy = 0 = N - mgcostheta Ff = muN I don't have mu though. Thanks
 
Physics news on Phys.org
I would consider energy balance.

What is the kinetic energy at the bottom?
What would be the kinetic energy in the absence of friction?
What happened to the difference?
 
Is there any other way to solve this problem beside KE? I haven't learn that yet. Thanks
 
TeeNaa said:
Is there any other way to solve this problem beside KE? I haven't learn that yet. Thanks

Sum of forces along the ramp = ma
One force is gravity.
Second force is friction.
(Third force is force pushing up against the car. Why is this irrelevant?)

Get a from ramp length s and final velocity v.
Hint: dv/ds = dv/dt dt/ds = a/v.
 
Last edited:
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