The coefficient of static friction between the floor of a truck and a box resting on it is 0.370. The truck is travelling at 80.1 km/h. What is the least distance in which the driver can stop the truck and ensure that the box does not slide?
Well, let me tell you a joke first. Why did the scarecrow win an award? Because he was outstanding in his field! Now, let's get back to your question.
To solve this problem, we need to use the equation:
\(d = \frac{v^2}{2\mu g}\)
Where:
- d is the distance needed to stop the truck,
- v is the initial velocity of the truck,
- \(\mu\) is the coefficient of static friction,
- g is the acceleration due to gravity.
Given that the coefficient of static friction (\(\mu\)) is 0.370 and the velocity (v) of the truck is 80.1 km/h, we need to convert the velocity to m/s and use the value of g, which is approximately 9.8 m/s².
So, let's calculate it step by step:
First, let's convert the velocity to m/s:
\(v = 80.1 \times \frac{1000}{3600}\) m/s
Next, let's substitute the values into the equation:
\(d = \frac{(80.1 \times \frac{1000}{3600})^2}{2 \times 0.370 \times 9.8}\) m
Let's calculate that:
\(d = \frac{(80.1 \times 1000/36)^2}{2 \times 0.370 \times 9.8}\) m
After crunching the numbers, the value of d is approximately 948.9 meters.
So, the least distance the driver needs to ensure that the box does not slide is approximately 948.9 meters.
Now, I hope this answer didn't drive you crazy!
To find the least distance in which the driver can stop the truck and ensure that the box does not slide, we need to consider two forces acting on the box: the force due to static friction and the force due to inertia.
The force due to static friction can be calculated using the formula:
\(F_{\text{{friction}}} = \mu_s \cdot N\)
where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force.
The normal force (\(N\)) can be calculated as the product of the mass of the box (\(m\)) and the acceleration due to gravity (\(g\)):
\(N = m \cdot g\)
where \(g\) is approximately 9.8 m/s².
The force due to inertia (\(F_{\text{{inertia}}}\)) can be calculated using Newton's second law of motion:
\(F_{\text{{inertia}}} = m \cdot a\)
where \(m\) is the mass of the box and \(a\) is the acceleration.
Since the box is not sliding, the force due to static friction is equal to the force due to inertia:
\(F_{\text{{friction}}} = F_{\text{{inertia}}}\)
Substituting the above equations, we get:
\(\mu_s \cdot N = m \cdot a\)
\(\mu_s \cdot m \cdot g = m \cdot a\)
Canceling out the mass (\(m\)) from both sides, we have:
\(\mu_s \cdot g = a\)
Given that the coefficient of static friction (\(\mu_s\) ) is 0.370 and \(g\) is 9.8 m/s², we can find the acceleration (\(a\)).
\(a = \mu_s \cdot g\)
\(a = 0.370 \cdot 9.8\) m/s²
Now, to find the distance required to stop the truck, we can use the kinematic equation:
\(v^2 = u^2 + 2a \cdot s\)
where \(v\) is the final velocity (0 m/s), \(u\) is the initial velocity (80.1 km/h), \(a\) is the acceleration, and \(s\) is the distance.
First, let's convert the initial velocity from km/h to m/s:
\(u = 80.1 \cdot \frac{{1000}}{{3600}}\) m/s
Now, substitute the values into the kinematic equation:
\(0 = \left(80.1 \cdot \frac{{1000}}{{3600}}\right)^2 + 2 \cdot (0.370 \cdot 9.8) \cdot s\)
Simplify the equation:
\(0 = \left(\frac{{80.1 \cdot 1000}}{{36}}\right)^2 + 2 \cdot (0.370 \cdot 9.8) \cdot s\)
Solve for \(s\) to find the least distance required to stop the truck.