# Algebra Problem

#### bigdreamer#1

10+ Year Member
15+ Year Member
Ok, for what ever reason, my brain is not working right.

This is from Kaplan's Physical Science Review book on chapter two of physics "Newtonian Mechanics"
M=mass of block A (15)
A=acceleration of block A (2.5)
mk=coefficient of kinetic friction (0.14)
g=acceleration of gravity (9.8)
T=tension of string

Solving for T, from:

MA=T- mkMg

then the book shows:

T=M(mkg+A)
=15(0.14(9.8)+2.5)
=58.1Newtons

My algebra must be way out of practice because I can't recall how each step of this occurred. Can someone break it down detailed step by step algebra for me, from MA=T-mkMg, I'm confused on what to do algebraically with the variable of "M" on each side so that you end up with the equation for T=M(mkg+A)

#### Drakensoul

##### Senior Member
7+ Year Member
15+ Year Member
Using the basic F=MA as the start:

F=MA
Forces acting, Tension and Friction in Opposing Directions, Tension leading since it is greater:
T-(mkMg)=MA
T=MA+mkMg
Factor out the common M
T=M(A+mkg)
Substitute given information
T=15[2.5+(0.14*9.8)]
Solve
T=58.1 Newtons

#### irie

##### royale with cheese
7+ Year Member
15+ Year Member
I didn't see anything that complicated on the mcat...well, its not complicated but it is time consuming.

M(mkg+A) = M(mkg) + M(A)

#### klonoff

##### Member
7+ Year Member
15+ Year Member
So you want to go from MA=T-mkMg to T=M(mkg+A)?

Add +mkMg to each side. Get MA + mkMg= T.

MA + mkMg = M(A+mkg) = T

#### bigdreamer#1

10+ Year Member
15+ Year Member
Thanks, just need to be sure.

This thread is more than 17 years old.

Your message may be considered spam for the following reasons: