Problem 33 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of \(350 \mathrm{~N}\). For the first \(15 \mathrm{~m}\) the floor is frictionless, and for the next \(15 \mathrm{~m}\) the coefficient of friction is \(0.25 .\) What is the final speed of the crate?

Step-by-Step Solution

Answer

The final speed of the crate is approximately 12.07 m/s.

Step 1: Calculate the acceleration on the frictionless surface

The total force exerted on the crate in the frictionless section is the horizontal pulling force because there is no opposing frictional force. Calculate the acceleration using Newton's second law, \( F = ma \).The net force \( F \) is \( 350 \text{ N} \) and the mass \( m \) is \( 96 \text{ kg} \).\[ a = \frac{F}{m} = \frac{350}{96} \approx 3.65 \, \text{m/s}^2 \].

Step 2: Calculate the velocity after first 15 m

Use the kinematic equation to find the velocity after the crate travels across the frictionless section:\[ v^2 = u^2 + 2as \]where \( u = 0 \text{ m/s} \) (initial velocity), \( a = 3.65 \text{ m/s}^2 \), and \( s = 15 \text{ m} \).\[ v^2 = 0 + 2 \times 3.65 \times 15 \]\[ v = \sqrt{2 \times 3.65 \times 15} \approx 10.48 \text{ m/s} \].

Step 3: Calculate the acceleration on the frictional surface

For the second segment, the acceleration is affected by friction. Calculate the frictional force using \( f = \mu mg \), where \( \mu = 0.25 \).\[ f = 0.25 \times 96 \times 9.8 = 235.2 \text{ N} \]The net force is now \( 350 \text{ N} - 235.2 \text{ N} = 114.8 \text{ N} \). Calculate the new acceleration:\[ a = \frac{114.8}{96} \approx 1.197 \, \text{m/s}^2 \].

Step 4: Calculate the final velocity after the second 15 m

Use the kinematic equation again for the second segment to calculate the final velocity:\[ v_f^2 = v^2 + 2as \]where \( v = 10.48 \text{ m/s} \) is the initial velocity from the first segment, \( a = 1.197 \text{ m/s}^2 \), and \( s = 15 \text{ m} \).\[ v_f^2 = (10.48)^2 + 2 \times 1.197 \times 15 \]\[ v_f \approx \sqrt{109.8 + 35.91} \approx \sqrt{145.71} \approx 12.07 \text{ m/s} \].

Key Concepts

Kinematics

Kinematics is a branch of physics that focuses on the motion of objects without considering the causes of that motion. In any kinematics problem, various equations relate the quantities of displacement, initial velocity, final velocity, acceleration, and time. 
A common kinematic equation used in our problem relates these quantities as follows: 

\[v^2 = u^2 + 2as\]
where:

• \(v\) is the final velocity
• \(u\) is the initial velocity
• \(a\) is the acceleration
• \(s\) is the displacement

This equation is extremely useful when an object’s motion is affected by constant acceleration. In the exercise, the crate's motion begins with an initial velocity of 0 m/s on a frictionless surface, allowing us to easily compute the final velocity after traveling 15 m with constant acceleration. Using the calculations, we find that the final velocity for this initial segment is about 10.48 m/s.

Frictional Force

Frictional force is the resistance force that acts between the surfaces of two objects moving against each other. In real-world applications, friction plays a crucial role by opposing the motion and affecting the net force acting on the object.
The magnitude of the frictional force can be calculated using the formula: 

\[ f = \mu mg \]
where:

• \(f\) is the frictional force
• \(\mu\) is the coefficient of friction
• \(m\) is the mass of the object
• \(g\) is the acceleration due to gravity, approximately 9.8 m/s²

When the crate crosses onto the section with a coefficient of friction of 0.25, the frictional force becomes a new factor. It reduces the net force acting on the crate, impacting its acceleration. In our problem, the frictional force is determined to be 235.2 N, which reduces the original pulling force, hence affecting the acceleration of the crate over the next 15 m.

Net Force

Net force is the vector sum of all forces acting upon an object. According to Newton’s Second Law, the net force enables us to determine the acceleration of an object through the equation: 

\[ F = ma \]
where:

• \(F\) is the net force
• \(m\) is the mass of the object
• \(a\) is the acceleration

Initially, when there is no friction, the entire pulling force of 350 N acts as the net force. Once the crate reaches the section of the floor with friction, the net force changes due to the opposing frictional force. The net force then becomes the applied force minus the frictional force, that is 350 N - 235.2 N = 114.8 N.
This new net force leads to a re-calculated acceleration for the second phase of the crate's movement, maintaining a clear relationship between forces acting on the crate and its subsequent motion.