Answer:
s = 0.2203 miles
Step-by-step explanation:
Given:
- The initial speed vo = 48 mi/hr
- The mass of the car m
- The coefficient of kinetic friction uk = 0.61
- The slope of the road 4-percent grade
Find:
Determine the stopping distance sAB. Repeat your calculations for the case when the car is moving downhill from B to A.
Solution:
- Apply the energy principle with work done against friction:
K.E_1 + P.E_1 = Wf + P_E_2 + K.E_2
- The final velocity of car is zero, K.E_2 = 0
The initial potential energy is set as zero, P.E_1 = 0
K.E_1 = Wf + P_E_2
0.5*m*vo^2 = Ff*s + m*g*h
Where, Ff is the frictional force:
Ff = uk*N
Where, N is the normal contact force between car and road. By equilibrium equation we have:
m*g*cos(θ) - N = 0
N = m*g*cos(θ)
Hence,
Ff = uk*m*g*cos(θ)
- The vertical distance travelled h is:
h = s*sin(θ)
- The energy equation is:
0.5*m*vo^2 = uk*m*g*cos(θ)*s + m*g*s*sin(θ)
0.5*vo^2 = uk*g*cos(θ)*s + g*s*sin(θ)
s*(uk*g*cos(θ) + g*sin(θ) ) = 0.5*vo^2
s = [0.5*vo^2 / (uk*g*cos(θ) + g*sin(θ) ) ]
- The slope = 4 / 100,
s = [0.5*48^2 / (0.61*8052.97*cos(2.29) + 8052.97*sin(2.29) ) ]
s = 0.2203 miles
