Constant frictional resistance & retardation

[es4]

Homework Statement

i) A car of mass 1.15 tonne, travelling on a straight level road, steadily rolls to rest from a speed of 20 m/s, in a distance of 800 m.

Determine, to 1 decimal place, the constant frictional resistance that causes the retardation.

What the question mean by "constant frictional resistance" in terms of formula?

Relevant Equations

f = m *a
f (friction) = mu * N (weight m*g)

What the question mean by "constant frictional resistance" in terms of formula?

I suppose they meant... "resistance force (f)" which is, >> f = m * a. So, here is my try...

m = 1.15 ton or 1150 kg
vi = 20 m/s
vf = 0
d = 800 m

v^2 - u^2 = 2ad
0 - 20^2 = 2a* 800
- 400 = 2a* 800
- 400 / 800 = 2a
- 0.5 = 2a
- 0.5 / 2 = a
- 0.25 or -0.3 = a m/s^2

constant friction force….
f = mass * acceleration
f = 1150 * - 0.25
f = -287.5 N <<<ans.?

Or, I am wrong? Does "constant frictional resistance" means, the value of "mu" or maybe "f (friction)" which is, >> f (friction) = mu * N (weight m*g)?

 

[Lnewqban]

I believe that the problem asks for the value of a constant horizontal force which is applied in a direction opposite to the movement of the car.

 

[es4]

I believe that the problem asks for the value of a constant horizontal force which is applied in a direction opposite to the movement of the car.

Do you mean, just the rate of deceleration? which was >>> a = - 0.25 or -0.3 m/s^2

 

[Lnewqban]

No, just imagine replacing the lady in the picture with the car of our problem.
The dragging tire induces a constant friction force that opposes the movement.

 

[es4]

No, just imagine replacing the lady in the picture with the car of our problem.
The dragging tire induces a constant friction force that opposes the movement.

Ok, that means I suppose to find "f (friction)" <<<<< which is, f (friction) = mu * N (weight m*g)... thanks.

 

[Lnewqban]

Ok, that means I suppose to find "f (friction)" <<<<< which is, f (friction) = mu * N (weight m*g)... thanks.

The cause is friction, the efect is decelerating force.
We could have many combinations of N and mu resulting in the same force opposing the movement.

A heavier tire on smooth wet concrete could produce similar dragging force or "frictional resistance (to linear forward movement)" than a lighter tire on rough dry asphalt.

Please, see this excellent tutorial on friction applied to several cases:
https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/

:)

 

[haruspex]

which is, f (friction) = mu * N (weight m*g)

Not quite.
The term "frictional resistance" here refers to what I prefer to call "rolling resistance". It is a combination of forces that act to impede motion even though the wheel is in rolling contact. It includes friction in the axle and losses due to deformation, but not friction between wheel and road surface, neither static nor kinetic.
See the link @Lnewqban provided.