The coefficiant of friction between rubber tires and wet pavement is 0.5. Breaks are alllied to a 750kg car, intitally travelling at 30 m/s and the car skids to a stop. What is the magnitude and direction of the acceleration. Ik the magnitude and i kinda get how to do the question, but i dont understand why we have to use force of friction as Fnet
M*g = 750 * 9.8 = 7350 N. = Wt. of car = Normal force(Fn).
Fp = 7350*sin 0o = 0 = Force parallel to the plane.
Fk = u*Fn = 0.5 * 7350 = 3675 N. = Force of kinetic friction.
Fp-Fk = M*a.
0-3675 = 750a, a = -4.9 m/s^2.
Or
u = a/g.
0.5 = a/-9.8, a = -4.9 m/s^2.
To determine the magnitude and direction of the acceleration, we need to consider the forces acting on the car. In this case, the primary force causing the car to decelerate is the force of friction between the tires and the wet pavement.
The force of friction can be calculated using the equation:
Frictional force (Ff) = coefficient of friction (μ) × normal force (Fn)
Where:
- μ is the coefficient of friction between the tires and the wet pavement, given as 0.5.
- Fn is the normal force acting on the car, which is equal to the weight of the car (mass × gravitational acceleration). In this case, the weight of the car is 750 kg × 9.8 m/s^2.
Since the car is skidding to a stop, the force of friction provides the net force acting on the car, accelerating it in the opposite direction of motion. Therefore, we can use the equation:
Net force (Fnet) = mass (m) × acceleration (a)
We need to find the acceleration. Rearranging the equation, we have:
a = Fnet / m
Since we know the mass of the car is 750 kg, we can substitute that value in. We also know that the net force is equal to the force of friction:
a = Ff / m
Now we can calculate the net force:
Ff = μ × Fn
Substituting the known values:
Ff = 0.5 × (750 kg × 9.8 m/s^2)
Once we calculate Ff, we can substitute it back into the acceleration equation to find the magnitude and direction of the acceleration.