aterials/gp/3411297783RCESGRAWE REPORT01otion with friction practice.docx1 /2 >157% e - Iab]1 (sha 25 box acrss w at constet seeedfHoe much fore do vau exert the baxMK-20.12. klween yo5 talle and nendis othbhat is gretesr hori2 fore eveyed whie t rerins2.Ms1s D4710crateaccelesation of

Answered: 1. Posh a 25 box across floor at…

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter6: Applications Of Newton's Laws

Section: Chapter Questions

Problem 46P: Friction (a) When rebuilding his car’s engine, a physics major must exert 3.00102Nof force to insert...

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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

Expert Solution

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.

The constant velocity implies zero acceleration.

Step 2

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

The net force equation for vertical motion.

$N - m g = F_{n e t, y} N - m g = 0 N = m g$

The net force equation for horizontal motion.

$\begin{array}{rcl} F - F_{k} & = & F_{n e t, x} \\ F - F_{k} & = & m a_{x} \\ F - F_{k} & = & m \times (0) \\ F & = & F_{k} \\ F & = & μ_{k} N \\ = & μ_{k} \times (m g) \\ = & 0.2 \times (25 k g \times 9.8 m / s^{2}) \\ F & = & 49 N \end{array}$

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

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