Question

An engine block of mass \(M\) is on the flatbed of a pickup truck that is traveling in a straight line down a level road with an initial speed of \(30.0 \mathrm{~m} / \mathrm{s}\). The coefficient of static friction between the block and the bed is \(\mu_{s}=0.540 .\) Find the minimum distance in which the truck can come to a stop without the engine block sliding toward the cab.

Short answer

Answer: The minimum stopping distance for the truck, without the engine block sliding forward, is approximately 85.2 meters.

Step-by-step solution

Analyze the forces acting on the block

The only horizontal force acting on the block is static friction since the truck is moving only in the horizontal direction. According to Newton's second law of motion, the force acting on the block (static friction) is equal to the mass of the block times its acceleration (or deceleration in this case). As the static friction force is what keeps the block from sliding, it can be written as: \(f_s = \mu_s F_N = \mu_s M g\) where \(f_s\) is the static friction, \(\mu_s\) is the coefficient of static friction, \(F_N\) is the normal force, \(g\) is the acceleration due to gravity, and \(M\) is the mass of the engine block.

Calculate the maximum possible deceleration

Using Newton's second law of motion for the horizontal motion of the block, we have: \(f_s = M a_{max}\) Substituting \(f_s\) from step 1, we get: \(\mu_s M g = M a_{max}\) Solve for the maximum possible deceleration (\(a_{max}\)) before the block starts to slide: \(a_{max} = \mu_s g\)

Calculate the minimum stopping distance

We will use the equation of motion to calculate the minimum stopping distance (\(d_{min}\)) of the truck without causing the engine block to slide forward: \(v^2 = u^2 + 2 a d\) Where \(v\) is the final speed (which should be 0), \(u\) is the initial speed, \(a\) is the acceleration (deceleration in this case, which is \(-a_{max}\)) and \(d\) is the stopping distance. Now, we plug in the values for final and initial speeds, and the deceleration: \(0 = 30.0^2 + 2 (-a_{max}) d_{min}\) Solve for \(d_{min}\): \(d_{min} = \frac{30.0^2}{2 a_{max}}\)

Calculate the values of deceleration and minimum stopping distance

Now we can plug in the values we have and calculate the minimum stopping distance: First, we calculate the maximum deceleration using the coefficient of static friction and gravitational acceleration: \(a_{max} = (0.540)(9.8) = 5.292 \mathrm{~m/s^2}\) Then, we plug in the maximum deceleration into the stopping distance formula: \(d_{min} = \frac{30.0^2}{2(5.292)} \approx 85.2 \mathrm{~m}\) The minimum stopping distance for the truck, without the engine block sliding forward, is approximately 85.2 meters.

Key concepts

Newton's second law of motion

Understanding Newton's second law of motion is crucial while solving static friction physics problems. The law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In simple terms, it's expressed as the equation
\[ F = ma \]
where F is the net force, m is the mass, and a is the acceleration. During braking, the force that prevents the sliding of objects—like our engine block on the truck bed—is the static frictional force. More specifically, the deceleration, or negative acceleration, of the truck is provided by the static frictional force acting in the opposite direction to the truck's movement. In physics problems, we often need to calculate this force to understand motion dynamics, such as the minimum distance required to stop a moving vehicle without disrupting its load.

Coefficient of static friction

Static friction comes into play when two objects are in contact but not moving relative to each other. The coefficient of static friction, denoted by \( \mu_s \), is a dimensionless number that describes how strongly the two surfaces grip each other. It varies depending on the materials of the surfaces. To calculate the static frictional force, you can use the formula:
\[ f_s = \mu_s F_N \]
where f_s is the static frictional force, and F_N is the normal force—basically, how hard the surfaces push against each other, which is typically the weight of an object (\(mg\)) in many basic scenarios. For the truck and engine block scenario, we factor in the coefficient to ensure the block doesn't slide as the truck comes to a stop.

Deceleration calculations

Deceleration refers to a decrease in velocity over time, which is acceleration in a direction opposite to the velocity of the object. In our textbook problem, calculating the deceleration is critical to determining how quickly the truck can stop without the engine block sliding.

We can deduce the maximum possible deceleration afforded by the static friction using the formula:
\[ a_{max} = \mu_s g \]
Once we have the maximum deceleration value, it can be used to find out various other motion parameters, such as stopping distance. In real-world applications, calculating deceleration is essential for the safe design of braking systems for vehicles.

Minimum stopping distance

The minimum stopping distance is the shortest distance over which a vehicle can be brought to a complete stop without causing any unsecured items to slide or fall over. To find this, we use the motion equation:
\[ v^2 = u^2 + 2ad \]
In a context where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration (or deceleration, and hence it would be negative in braking scenarios), and 'd' is the stopping distance. This is an important calculation for ensuring road safety as it helps in defining speed limits and constructing appropriate warning systems before stops or sharp bends. By carefully considering the minimum stopping distance, drivers can avoid collisions and transport items more securely.