**Physics Problem: Block on a Rough Surface****Problem Statement:**A 5 kg block is pulled along a rough, level surface by a force \( P \), where \( P = 300 \, \text{N} \). The figure illustrates the free-body diagram for the block. \( f_k \) represents the force of kinetic friction between the block and the surface, with a coefficient of friction \( \mu_k = 0.3 \). Calculate the magnitude of the block's acceleration.**Hints:**- Use two equations from Newton’s second law. Only one directly involves acceleration, but both are needed. **Diagram Explanation:**- A block labeled 5 kg is on a horizontal surface.- \( f_k = 0.3 \) denotes the force of kinetic friction opposing the motion.- \( P = 300 \, \text{N} \) is the horizontal force pulling the block.- \( F_N \) is the normal force acting perpendicular to the surface.- \( mg \), the weight of the block, acts downward.**Equations:**1. Sum of forces in the x-direction: \[ \Sigma F_x = ma_x = P - \mu_k F_N \]2. Sum of forces in the y-direction (equilibrium, no vertical acceleration): \[ \Sigma F_y = 0 = F_N - mg \]**Solution Approach:**- Use the second equation to solve for the normal force \( F_N \): \[ F_N = mg \]- Plug \( F_N \) into the first equation to find the acceleration \( a \): \[ P - \mu_k mg = ma \]**Options for Acceleration:**- A) 25 m/s\(^2\)- B) 57 m/s\(^2\)- C) 285 m/s\(^2\)- D) 315 m/s\(^2\)- E) None of the aboveUsing this setup will guide you through solving for the acceleration of the block on this rough surface.

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20. A 5 kg block is pulled along a rough level surface by a pull force P, where P = 300 N. The figure shows the free-body diagram for the block, fk represents the force of kinetic friction. If the coefficient of kinetic friction between the block and the surface, k = 0.3, calculate the magnitude of the block's acceleration? Hint: Obtain two equations from Newton's second law, only one of them directly involve acceleration but you will still need the other equation. FN fko.? 3 5kg mg 300N olay ontos antiga P ΣFx = max= P_-4₁₂² = Max {Fy=may = FN-pg = mal FN-g=a el or sups P-MR₂ = m (FN Мк

20. A 5 kg block is pulled along a rough level surface by a pull force P, where P = 300 N. The figure shows the free-body diagram for the block, fk represents the force of kinetic friction. If the coefficient of kinetic friction between the block and the surface, k = 0.3, calculate the magnitude of the block's acceleration? Hint: Obtain two equations from Newton's second law, only one of them directly involve acceleration but you will still need the other equation. FN fko.? 3 5kg mg 300N olay ontos antiga P ΣFx = max= P_-4₁₂² = Max {Fy=may = FN-pg = mal FN-g=a el or sups P-MR₂ = m (FN Мк

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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Subject: physics
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