6. A car is moving along a straight road at 60 mph [26.7 meter/sec ]. The driver makes slams on his breaks locking them so that the car skid s to a stop. Assume a constant braking force due to friction , and a kinetic friction coefficient of 0.70 between the tires and the road: (a) calculate the braking distance from the instant that the brakes were applied, (b0 how much work did the friction do in stopping the car?
Vo = 26.7 m/s
V = 0
u = 0.70
a. u = a/g
a = g*u = (-9.8) * 0.70 = -6.86 m/s^2.
V^2 = Vo^2 + 2a*d = 0 @ max distance.
d = -(Vo^2)/2a = -(26.7^2)/-13.72 = 52 m
How does this work
To calculate the braking distance, we need to first determine the deceleration (negative acceleration) of the car. We can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = ma).
We know that the braking force is due to friction, and the formula for frictional force is given by the equation F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. The normal force in this case is the same as the weight of the car, which is equal to the mass of the car multiplied by the acceleration due to gravity.
a) Calculating the deceleration:
Since the car is skidding, the maximum frictional force (F_friction) is equal to the maximum force applied by the brakes. So we can equate F = F_friction:
ma = μ * N
From this equation, we can solve for the acceleration, a:
a = μ * (m * g)
Given:
μ = 0.70 (coefficient of kinetic friction)
m = mass of the car (not given)
g = acceleration due to gravity (approximately 9.8 m/s^2)
b) Calculating the braking distance:
The braking distance (d) can be obtained using the SUVAT equation, which relates displacement, initial velocity, final velocity, acceleration, and time: v^2 = u^2 + 2as.
In this case, the final velocity (v) is 0 (since the car comes to a stop), the initial velocity (u) is 26.7 m/s, and the acceleration (a) is the deceleration we calculated earlier.
Now let's calculate each part of the question step by step:
a) Calculate the deceleration, a:
a = μ * (m * g) = 0.70 * (m * 9.8)
b) Calculate the braking distance, d:
0^2 = (26.7)^2 + 2 * a * d
Simplifying the equation: 0 = 712.89 + 19.6 * d
Rearranging the equation to solve for d:
19.6 * d = -712.89
d = -712.89 / 19.6
The braking distance from the instant the brakes were applied can be calculated as d ≈ 36.36 meters.
To find out the work done by friction, we can use the equation W = F * d, where W is work, F is force, and d is displacement.
b) Calculate the work done by friction:
The force of friction can be calculated using F_friction = μ * N, and since N is the weight of the car (m * g), we have F_friction = μ * (m * g).
The work done by friction is then:
W = F_friction * d
Now we can substitute the values:
W = (μ * (m * g)) * 36.36
Since the mass (m) of the car is not given, we cannot calculate the exact work done by friction without that information.