a 1000 kg truck traveling at a speed of 40m/s skids to a halt on wet concrete where µk = 0.60. how long are the skid marks?
(Please explain all variables and equations I'm clueless in this subject)
I keep trying to use the equation I found online: d= v^2/2ug
but I do not fully understand the variables, "u" and "g."
To solve the problem, you will be using the equation d = (v²) / (2μk * g), where:
- d represents the length of the skid marks,
- v represents the initial velocity of the truck,
- μk represents the coefficient of kinetic friction between the tires and the wet concrete,
- g represents the acceleration due to gravity (approximately 9.8 m/s²).
Let's break down the equation step by step:
1. v: In this problem, the initial velocity of the truck is given as 40 m/s.
2. μk: The coefficient of kinetic friction (μk) represents the force resisting the motion of the truck on the wet concrete. The given value is 0.60.
3. g: The acceleration due to gravity (g) represents the rate at which objects fall towards the Earth. The value is approximately 9.8 m/s². While it may seem unrelated to this question, it plays a role in calculating the frictional force.
Now, let's substitute the given values into the equation:
d = (v²) / (2μk * g)
d = (40²) / (2 * 0.60 * 9.8)
d = 1600 / (1.176)
d ≈ 1362.6 meters
Therefore, the length of the skid marks is approximately 1362.6 meters.
It's important to note that the coefficient of kinetic friction (μk) can vary depending on the surface conditions (dry, wet, icy, etc.) and the type of materials involved. This value might be provided in the problem statement or could be obtained from reference tables.