Problem with regards to coefficient of kinetic friction

Question

a 900kg car travelling at 25m/s skids to a stop on wet surface after the brakes are applied. how far does the car skid if the coefficient of kinetic friction is 0.30? show the steps in solving.

Answer 1

This is a simple question. Your car is traveling 25 m/s @ 900 kg And the question wants you to figure out how long it will take to stop with a coefficient of friction of .3 which is high for a wet surface but anyways. This can be solved simply(assuming car is not a slope but a flat surface otherwise that information changes this) :

m: 900 kg         Normal Force: 900 * 9.8 = 8829 N

V: 25 m/s         Friction Force:  Normal Force* Coefficient of Friction

                                              8829 * .3 = 2648.7 N

Now use momentum to figure out the time it stopped so that we can then figure out the distance it stopped you are given the mass and velocity which means you can use the formula

mass * velocity = force * time

since we figured out the force acting to slow the car down above.

900* 25 = 2648.7 * T

T = 8.49 s

now the car changes from 25 m/s to 0 m/s in 8.49s. This is how long it took to stop

therefore we can do the change in speed divided by the time to figure out the rate at which the car deccelerates.

25 m/s / 8.49 s = 2.943 m/ s^2

Now we are in our final step we have all the information we need for our formula to figure out distance

(Final Velocity) ^2 = (Initial Velocity)^2 + 2*acceleration*distance

(0)^2 = 25^2 + 2* 2.943*D

D = 106 m

Answer 2

hi

initial velocity = vi = 25 m/s

final velocity = vf = 0

mass = m = 900 kg

coefficient of kinetic friction = k = 0.3

distance = x m

As F = ma = -kx

=> ma = -kx   --------------(1)

a = ?

As 2ax = vf^2 - vi^2

=> 2ax = 0^2 - 25^2

=> 2ax = 0 - 625

=> a = -625/2x

=> a = -312.5/x

putting all values in (1)

=> ma = -kx   --------------(1)

=> 900(-312.5/x) = -0.3(x)

=> -281250/x = -0.3x

=> -281250/0.3 = x^2

=> 937500 = x^2

=> sqrt(937500) = x

=> 968.24 = x

Therefore the car skids for about 968.24 m. Hope to answer you well. Keep smiling. Bye.

Answer 3

The only normal force on the car is exerted by the surface. And as the car has no upward acceleration, N = mg

Friction  = umg = 0.3*9000 = 2700 N

Acceleration = Force/Mass

                   = 2700/900 = 3 m/s^2

V^2 - U^2 = 2aS

-625 = 6S

S = 104 m

Answer 4

Frictional force

= μmg

Deceleration

= Force/mass

= μmg/m

= μg

= 0.30 * 9.8

= 2.94 m/s^2

Initial velocity, u = 25 m/s

Final velocity, v = 0

Acceleration, a = - 2.94 m/s^2

Distance traveled, s = ?

v^2 - u^2 = 2as

=> s = u^2/2a

= (25)^2 / 2*2.94

= 106 m.