2000-kg car moving at an initial speed of 25 m/s along a horizontal road skids to a stop in 60 m. (a) Find the energy dissipated by friction (b) Find the coefficient of kinetic friction between the tires and the road_ (Note: When stopping without skidding and using conventional brakes; 100 percent of the kinetie energy is dissipated by friction within the brakes:
The Correct Answer and Explanation is:
We are given:
- Mass of car, m=2000 kgm = 2000 \, \text{kg}m=2000kg
- Initial speed, vi=25 m/sv_i = 25 \, \text{m/s}vi=25m/s
- Final speed, vf=0 m/sv_f = 0 \, \text{m/s}vf=0m/s
- Skid distance, d=60 md = 60 \, \text{m}d=60m
(a) Energy dissipated by friction
Since the car comes to a stop, all its initial kinetic energy is lost due to friction.
The initial kinetic energy (KE) is: KE=12mv2=12(2000)(25)2=1000×625=625,000 JKE = \frac{1}{2}mv^2 = \frac{1}{2}(2000)(25)^2 = 1000 \times 625 = 625{,}000 \, \text{J}KE=21mv2=21(2000)(25)2=1000×625=625,000J
So, the energy dissipated by friction is: 625,000 J\boxed{625{,}000 \, \text{J}}625,000J
(b) Coefficient of kinetic friction, μk\mu_kμk
To find the coefficient of kinetic friction, we use the work-energy principle. The work done by friction is equal to the loss of kinetic energy: W=fk⋅d=μk⋅N⋅d=μk⋅mg⋅dW = f_k \cdot d = \mu_k \cdot N \cdot d = \mu_k \cdot mg \cdot dW=fk⋅d=μk⋅N⋅d=μk⋅mg⋅d
Set this equal to the energy dissipated: μk⋅mg⋅d=12mv2\mu_k \cdot mg \cdot d = \frac{1}{2}mv^2μk⋅mg⋅d=21mv2
Cancel out mmm from both sides: μk⋅g⋅d=12v2\mu_k \cdot g \cdot d = \frac{1}{2}v^2μk⋅g⋅d=21v2
Solve for μk\mu_kμk: μk=v22gd=2522⋅9.8⋅60=6251176≈0.531\mu_k = \frac{v^2}{2gd} = \frac{25^2}{2 \cdot 9.8 \cdot 60} = \frac{625}{1176} \approx 0.531μk=2gdv2=2⋅9.8⋅60252=1176625≈0.531
So, the coefficient of kinetic friction is: 0.531\boxed{0.531}0.531
Explanation
When a car moves and then stops due to skidding, its kinetic energy is converted entirely into heat and sound through friction. This means that the energy the car had while moving is used up to overcome the force of friction between the tires and the road.
To calculate the energy lost, we use the formula for kinetic energy. This energy depends on the mass of the car and the square of its speed. Since the final speed is zero, the entire kinetic energy is lost during braking. Plugging in the values, we find that 625,000 joules of energy were dissipated as the car skidded to a stop.
Next, to find the coefficient of kinetic friction, we consider how much force was needed to stop the car over a given distance. The friction force is what slows the car down. This force is the product of the normal force and the coefficient of friction. Since the car is on a horizontal road, the normal force equals the weight of the car. Using the work-energy principle, we equate the work done by this friction force to the car’s lost kinetic energy. This allows us to solve for the coefficient.
The result, approximately 0.531, is reasonable for rubber tires on dry pavement. A higher value would suggest a more abrupt stop, and a lower value would indicate less effective friction. This coefficient is crucial in real-world applications such as designing road surfaces and braking systems to ensure vehicles can stop safely.
