1. Posh a 25 box across floor at constat speed of lmb. MK 0,2 How much force do you exert on the box?

University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter12: Static Equilibrium And Elasticity
Section: Chapter Questions
Problem 22CQ: Note: Unless stated othen.ise the weights of the wires, rods, and other elements are assumed to be...
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aterials/gp/3411297783
RCES
GRAWE REPORT
01
otion with friction practice.docx
1 /2 >
157% e - I
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1 (sha 25 box acrss w at constet seeedf
Hoe much fore do vau exert the bax
MK-2
0.1
2. klween yo5 talle and nendis oth
bhat is gretesr hori2 fore eveyed whie t rerins
2.Ms
1s D4
710
crate
accelesation of <rate
Transcribed Image Text:aterials/gp/3411297783 RCES GRAWE REPORT 01 otion with friction practice.docx 1 /2 > 157% e - I ab] 1 (sha 25 box acrss w at constet seeedf Hoe much fore do vau exert the bax MK-2 0.1 2. klween yo5 talle and nendis oth bhat is gretesr hori2 fore eveyed whie t rerins 2.Ms 1s D4 710 crate accelesation of <rate
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Step 1

Since you have asked multiple questions, we will solve the first question for you. If you want a specific question to be solved then please specify the question number or post only that question.

Physics homework question answer, step 1, image 1

 

The constant velocity implies zero acceleration.

Step 2

Let's write a net force equation in terms of frictional force Fk , the weight mg.

The net force equation for vertical motion.

N-mg=Fnet,yN-mg=0N=mg

The net force equation for horizontal motion.

F-Fk=Fnet,xF-Fk=maxF-Fk=m×(0)F=FkF=μkN=μk×mg=0.2×25 kg×9.8 m/s2F=49 N

Therefore, the force exerted on the box is 49N.

 

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