A car with mass of 1000kg travelling at speed of 108km/hr along level road is brought to stop uniformly in distance of 70m. What was the coefficient of friction between the tires and the the roads?
108 km/hr = 108,000 meters / 3600 seconds = 30 meters/seconds
d =70 = average speed * t = 15 t
so t = 70/15 = 14/3
d = Vi t + (1/2) a t^2
70 = 30 (14/3) + (1/2) a (14/3)^2
70 = 140+ 10.9 a
10.9 a = -70
a = - 6.43 m/s^2
F = m a = - 6430 Newtons
6430 = mu m g
so
mu = 6.430 / 9.81 = 0.655
To find the coefficient of friction between the tires and the road, we can use the equation:
\(F = \mu * N\)
Where:
- \(F\) is the force of friction
- \(\mu\) is the coefficient of friction
- \(N\) is the normal force
The normal force can be calculated as:
\(N = mg\)
Where:
- \(m\) is the mass of the car
- \(g\) is the acceleration due to gravity (9.8 m/s\(^2\))
First, let's convert the speed from km/hr to m/s. We know that \(1 \text{ km/hr} = \frac{1000}{3600} \text{ m/s}\). Therefore:
\(108 \text{ km/hr} = 108 \times \frac{1000}{3600} \text{ m/s} = 30 \text{ m/s}\)
To find the force of friction, we can use Newton's second law of motion:
\(F = ma\)
The acceleration (\(a\)) can be calculated using the formula:
\(a = \frac{{v^2}}{{2s}}\)
Where:
- \(v\) is the initial velocity of the car
- \(s\) is the distance over which the car is brought to a stop
Let's plug in the values:
\(a = \frac{{(30)^2}}{{2 \times 70}} \text{ m/s}^2\)
Now, we can find the force of friction:
\(F = ma = 1000 \times \frac{{(30)^2}}{{2 \times 70}} \text{ N}\)
Finally, we can find the coefficient of friction:
\(F = \mu \times N\)
Substituting our values:
\(1000 \times \frac{{(30)^2}}{{2 \times 70}} = \mu \times mg\)
Let's solve for \(\mu\):
\(\mu = \frac{{1000 \times \frac{{(30)^2}}{{2 \times 70}}}}{{mg}}\)
Substituting \(m = 1000 \, \text{kg}\), \(g = 9.8 \, \text{m/s}^2\):
\(\mu = \frac{{1000 \times \frac{{30^2}}{{2 \times 70}}}}{{1000 \times 9.8}}\)
To find the coefficient of friction between the tires and the road, we can use the equation for deceleration:
a = (v^2 - u^2) / (2s),
where
a is the deceleration,
v is the final velocity (0 m/s since the car is brought to a stop),
u is the initial velocity (108 km/hr converted to m/s),
s is the distance covered (70 m).
First, let's convert 108 km/hr to m/s:
108 km/hr * (1000 m/1 km) * (1 hr / 3600 s) = 30 m/s.
Now, we can substitute the values into the equation:
a = (0^2 - 30^2) / (2 * 70).
Calculating the numerator:
0^2 - 30^2 = -900.
Calculating the denominator:
(2 * 70) = 140.
Now, substitute the values back into the equation:
a = -900 / 140.
Calculating the deceleration:
a ≈ -6.43 m/s^2.
Since the car is brought to a stop in a uniform manner, the deceleration is equal to the coefficient of friction multiplied by the acceleration due to gravity. Therefore, we can write:
a = μ * g,
where
μ is the coefficient of friction,
g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation for μ:
μ = a / g.
Substituting the values:
μ = -6.43 / 9.8 ≈ -0.656.
The coefficient of friction between the tires and the road is approximately -0.656. It is important to note that the negative sign indicates the direction of the frictional force, opposite to the direction of motion of the car.