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(II) A person pushes a 14.0-kg lawn mower at constant

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 26P Chapter 4

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 26P

Problem

(II) A person pushes a 14.0-kg lawn mower at constant speed with a force of F =88.0 directed along the handle, which is at an angle of 45.0" to the horizontal (Fig. 4-58). (a) Draw the free-body diagram showing all forces acting on the mower. Calculate (h) the horizontal friction force on the mower, then (c) the normal force exerted vertically upward on the mower by the ground, (a) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force?

Step-by-Step Solution:

Solution 26P: We have to calculate the forces exerted on lawn mower and the forces exerted by the lawn mower under the given situation. Step 1 of 5 Concept: 1) Newton’s third law: To every action there is an equal and opposite reaction. 2) Newton’s second law: The force F acts on mass m produces an acceleration a in the object. Mathematically, F=ma. 3) Frictional force also acts on the surface of the bodies in contact whenever their is relative motion between between the two bodies. 4) Free body diagram of a body gives the diagrammatical representation of all the forces acting on the body in terms of magnitude and diagram, under the given situation. Step 2 of 5 A] Free body diagram of the lawn mower is shown below. o Where, F Force applied to lawn-mower at angle of = 45 Fcos Horizontal component of applied force F Fsin Vertical component of applied force F mg Weight of the lawn-mower acting verting downward F N Normal forces exerted by the ground on the lawn-mower F frFrictional force between ground and lawn-mower Step 3 of 5 B] Frictional force acting on the lawn-mower Mass of the lawn mower m = 14 kg Force applied by person F = 88 N As the body is moving with constant speed along horizontal direction, therefore, using Newton’s second law, we get, F x Fcos F = ma = m×0 = 0 fr F = Fcos fr F = 88×cos45 o fr F = 62.2 N . fr

Step 4 of 5

Chapter 4, Problem 26P is Solved
Step 5 of 5

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

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(II) A person pushes a 14.0-kg lawn mower at constant

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