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Magazine Chapter 6: Problem 73
A 36.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m the floor is frictionless, and for the next 10.0 m the coefficient of friction is 0.20. What is the final speed of the crate after being pulled these 21.0 m?
Short Answer
Expert verified
The final speed of the crate after 21.0 m is approximately 14.94 m/s.
Step by step solution
01
Understanding the Problem
We have a crate with an initial mass of 36.0 kg which starts from rest and is pulled with a force of 225 N. The crate first moves over a frictionless surface for 11.0 m and then over a surface with friction for the next 10.0 m. Our task is to find the crate's final speed after traveling a total of 21.0 m.
02
Calculate Acceleration and Velocity on Frictionless Surface
Since the surface is frictionless, the only force acting on the crate is the pulling force. Using Newton's second law, acceleration is given by \( a_1 = \frac{F}{m} = \frac{225}{36.0} \approx 6.25 \, \text{m/s}^2 \). The velocity at 11.0 m when the crate reaches the end of the frictionless surface can be calculated using the equation \( v_1^2 = u^2 + 2a_1s_1 \), where \( u = 0 \), \( a_1 = 6.25 \, \text{m/s}^2 \), and \( s_1 = 11.0 \, \text{m} \), leading to \( v_1^2 = 2 \times 6.25 \times 11.0 \). Solve for \( v_1 \).
03
Substitute and Calculate Velocity for Frictionless Surface
The calculation becomes \( v_1^2 = 137.5 \), so \( v_1 = \sqrt{137.5} \approx 11.72 \, \text{m/s} \). This is the velocity of the crate at the end of the first 11.0 meters.
04
Calculate Deceleration Due to Friction
For the next 10.0 m, the floor has a friction coefficient of 0.20. The frictional force is \( f = \mu \cdot m \cdot g = 0.20 \times 36.0 \times 9.81 \approx 70.56 \, \text{N} \). The net force is \( F - f = 225 - 70.56 = 154.44 \, \text{N} \).The deceleration due to friction is \( a_2 = \frac{154.44}{36.0} \approx 4.29 \, \text{m/s}^2 \).
05
Calculate Final Velocity after 21.0 m
Using the velocity from the end of the frictionless section as the initial velocity for the next section, we calculate the final velocity using \( v_2^2 = v_1^2 + 2a_2s_2 \), where \( s_2 = 10.0 \, \text{m} \), and \( v_1 \approx 11.72 \, \text{m/s} \). Plug in \( v_1 = 11.72 \, \text{m/s} \) and solve for \( v_2 \).
06
Substitute and Calculate for Final Velocity
Substitute into the equation: \( v_2^2 = 11.72^2 + 2 \times 4.29 \times 10 = 137.48 + 85.8 = 223.28 \). Hence, \( v_2 = \sqrt{223.28} \approx 14.94 \, \text{m/s} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Physics Problem Solving
Physics problem solving involves using a systematic approach to analyze and solve problems. It requires understanding the given information, determining the laws and equations applicable to the scenario, and following steps to arrive at a solution.
- **Understand the Problem**: Begin by clearly identifying what is given and what needs to be found. In our exercise, we must determine the final speed of a crate being pulled.- **Apply the Right Concepts**: Recognize that Newton's second law is crucial here, where force equals mass times acceleration (\(F = ma\)). This aids in calculating acceleration and velocity.- **Step-By-Step Calculation**: Follow a methodical approach of dividing the problem into smaller sections, like evaluating each section of movement鈥攆rictionless and with friction individually.
Such structured steps not only simplify the problem but also help prevent mistakes by promoting clear, logical progression towards the solution.
- **Understand the Problem**: Begin by clearly identifying what is given and what needs to be found. In our exercise, we must determine the final speed of a crate being pulled.- **Apply the Right Concepts**: Recognize that Newton's second law is crucial here, where force equals mass times acceleration (\(F = ma\)). This aids in calculating acceleration and velocity.- **Step-By-Step Calculation**: Follow a methodical approach of dividing the problem into smaller sections, like evaluating each section of movement鈥攆rictionless and with friction individually.
Such structured steps not only simplify the problem but also help prevent mistakes by promoting clear, logical progression towards the solution.
Kinematics
Kinematics is the study of motion without considering the forces causing it. It's closely related to Newton's laws as it uses equations to describe motion in terms of displacement, velocity, and acceleration.
In our exercise, we incorporate kinematic equations to find the velocity of the crate:
In our exercise, we incorporate kinematic equations to find the velocity of the crate:
- For the frictionless surface: Using the equation \(v^2 = u^2 + 2as\), where \(u\) is initial velocity (0 in this case), \(a\) is acceleration, and \(s\) is distance, we calculate the velocity at the end of the frictionless path.
- For movement with friction: Consider the ending speed of the frictionless section as the starting speed of this section, using the same equation form to find the final velocity after the friction part.
Friction and Motion
Friction is a force that opposes the relative motion of two surfaces in contact. It is crucial in physics to understand how it affects the motion of objects.
In our scenario:
In our scenario:
- **Frictional Force**: Calculated as \(f = \mu \times m \times g\), where \(\mu\) is the coefficient of friction, \(m\) is mass, and \(g\) is gravitational acceleration. This force acts in the opposite direction to the motion.
- **Impact on Motion**: On the frictionless surface, only the pulling force acts, leading to straightforward acceleration. When reaching the frictional surface, the net force is reduced by the frictional force, resulting in slower acceleration (deceleration actually).
- **Calculating with Friction**: Incorporate these forces into Newton's laws to adjust acceleration and ultimately calculate speed.
