A 2000 {\rm kg} car traveling at a speed of 13 {\rm m/s} skids to a halt on wet concrete where mu_k = 0.50. How long are the skid marks?

I tried doing this many times and each answer is wrong. Can someone help?

Fc = mg = 2kg * 9.8N/kg = 19.6N = Force

of car.

Fc = 19.8N @ 0 deg.(Hor. plane).

Fp = mg*sin(0) = 19.8sin(0) = 0 = Force
parallel with plane.

Fv = 19.6cos(0) = 19.6N = Force perpendicular to plane.

Ff = u*Fv = 0.5 * 19.6 = 9.8N.

Fn = Fp - Ff = 0 - 9.8 = -9.8N = Net
force.

Fn = ma,
a = Fn/m = -9.8/2 = -4.9m/s^2.

Vf^2 = Vo^2 + 2ad,
d = (Vf^2-Vo^2) / 2a,
d = (0-(13)^2 / -9.8 = 17.2m. = Length
of skid marks.

Correction:

mass of the car = 2000kg: NOT 2kg.

Fc = 2000kg * 9.8N/kg = 19,600N.

Fp = 19,600sin(0) = 0.

Fv = 19,600cos(0) = 19,600N.

Ff = u*Fv = 0.5 * 19,600 = 9800N.

Fn = Fp - Ff = o - 9800 = -9800N.

a = Fn/m = -9800 / 2000 = -4.9m/s^2.

The remaining calculations are correct
as is.

Thank you very much!

Glad i could help!

work = change in kinetic energy

friction force X distance = change in KE
-mu_k n x = 0 - 1/2 m v^2
[Remember, friction does negative work)
mu_k m g x = 1/2 m v^2
Simplifying:
x = v^2/(2 mu_k g) = 17.24 m

Notice that in this solution, the mass of the car is not needed.

To find the length of the skid marks, we can use the equation of motion for constant acceleration. However, since the car is skidding to a halt, its acceleration will be in the opposite direction of motion.

The equation of motion in this case is:
v^2 = u^2 + 2as

where:
v = final velocity (0 m/s, since the car has skidded to a halt)
u = initial velocity (13 m/s, given)
a = acceleration
s = distance or skid marks (what we need to find)

But first, we need to find the acceleration. In this case, the force of friction between the car's tires and the wet concrete is what causes the deceleration. The magnitude of the frictional force is given by the formula:

F_friction = μ_k * N

where:
μ_k = coefficient of kinetic friction (given as 0.50)
N = normal force

In this case, the normal force is equal to the weight of the car, which can be calculated as:

N = mg

where:
m = mass of the car (2000 kg, given)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Now, we can calculate the force of friction:

F_friction = μ_k * N
= μ_k * mg

Next, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration:

Net Force = ma

In this case, the net force is the force of friction, so:

F_friction = ma

Substituting the value of F_friction:

μ_k * mg = ma

We can now solve for the acceleration a:

a = μ_k * g

Substituting the values:

a = 0.50 * 9.8 m/s^2
a = 4.9 m/s^2

Now, let's plug this acceleration value into the equation of motion:

v^2 = u^2 + 2as

Rearranging the equation to solve for s:

s = (v^2 - u^2) / (2a)

Substituting the values:

s = (0^2 - 13^2) / (2 * -4.9 m/s^2)
s = (169) / (-9.8 m/s^2)
s ≈ -17.24 m

Since we're dealing with distance and not a vector, the negative sign can be ignored. So, the length of the skid marks is approximately 17.24 meters.