### Calculating the Pulling Force on a SledA 45.0-kg sled is pulled across a horizontal icy surface as shown in the figure. The horizontal pulling force of magnitude \(P\) is directed opposite to a constant 25.0-N kinetic friction force. Starting from rest, the sled acquires a final speed of 2.25 m/s after being pulled a distance of 28.0 m.**Given:**- Mass of the sled (\(m\)): 45.0 kg- Kinetic friction force (\(f_k\)): 25.0 N- Final speed (\(v_f\)): 2.25 m/s- Distance (\(d\)): 28.0 m**Diagram Explanation:**The provided figure shows a sled located on a flat, icy surface. An arrow labeled \(P\) points to the right, indicating the direction of the pulling force. Another arrow labeled \(f_k\) points to the left, indicating the direction of the kinetic friction force opposing the movement.**Problem:**What is the magnitude \(P\) of the pulling force?**Calculation:**To calculate the magnitude of the pulling force \(P\), we can use the work-energy principle and Newton's second law of motion.We start with the work-energy principle:The work done by all forces is equal to the change in kinetic energy.\[ W_{\text{total}} = \Delta KE = \frac{1}{2} m v_f^2\]Since the sled starts from rest, the initial kinetic energy is zero. Therefore, the kinetic energy change is:\[ \Delta KE = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 45.0\,\text{kg} \times (2.25\,\text{m/s})^2 \]Let's calculate this:\[ \Delta KE = \frac{1}{2} \times 45.0\,\text{kg} \times 5.0625 \,\text{(m/s)}^2 \]\[ \Delta KE = 113.90625 \,\text{J} \]The total work done is also expressed as \(W_{\text{total}} = (P - f_k) \times d\):\[ 113.90625 \,\text

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A 45.0-kg sled is pulled across a horizontal icy surface as in the figure. The horizontal pulling force of magnitude P is directed opposite to a constant 25.0-N kinetic friction force. Starting from rest, the sled acquires a final speed of 2.25 m/s after being pulled a distance of 28.0 m. P What is the magnitude P of the pulling force? N twikke Ⓡ

A 45.0-kg sled is pulled across a horizontal icy surface as in the figure. The horizontal pulling force of magnitude P is directed opposite to a constant 25.0-N kinetic friction force. Starting from rest, the sled acquires a final speed of 2.25 m/s after being pulled a distance of 28.0 m. P What is the magnitude P of the pulling force? N twikke Ⓡ

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Static and Kinetic Friction

Kinetic friction comes into existence when there are moving surfaces means in other words we can say that the force which comes into existing between moving surfaces is known as kinetic friction.

Question

### Calculating the Pulling Force on a Sled

A 45.0-kg sled is pulled across a horizontal icy surface as shown in the figure. The horizontal pulling force of magnitude \(P\) is directed opposite to a constant 25.0-N kinetic friction force. Starting from rest, the sled acquires a final speed of 2.25 m/s after being pulled a distance of 28.0 m.

**Given:**
- Mass of the sled (\(m\)): 45.0 kg
- Kinetic friction force (\(f_k\)): 25.0 N
- Final speed (\(v_f\)): 2.25 m/s
- Distance (\(d\)): 28.0 m

**Diagram Explanation:**
The provided figure shows a sled located on a flat, icy surface. An arrow labeled \(P\) points to the right, indicating the direction of the pulling force. Another arrow labeled \(f_k\) points to the left, indicating the direction of the kinetic friction force opposing the movement.

**Problem:**
What is the magnitude \(P\) of the pulling force?

**Calculation:**
To calculate the magnitude of the pulling force \(P\), we can use the work-energy principle and Newton's second law of motion.

We start with the work-energy principle:
The work done by all forces is equal to the change in kinetic energy.

\[
W_{\text{total}} = \Delta KE = \frac{1}{2} m v_f^2
\]

Since the sled starts from rest, the initial kinetic energy is zero. Therefore, the kinetic energy change is:
\[
\Delta KE = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 45.0\,\text{kg} \times (2.25\,\text{m/s})^2
\]

Let's calculate this:
\[
\Delta KE = \frac{1}{2} \times 45.0\,\text{kg} \times 5.0625 \,\text{(m/s)}^2
\]

\[
\Delta KE = 113.90625 \,\text{J}
\]

The total work done is also expressed as \(W_{\text{total}} = (P - f_k) \times d\):
\[
113.90625 \,\text

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