I'm stuck Not sure where to go. Help

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The discussion revolves around calculating the acceleration of a two-mass system involving friction. The setup includes a 10 kg mass (m1) on a table connected to a 4 kg hanging mass (m2) with specified coefficients of static and kinetic friction. The user initially misinterpreted the problem as involving a frictionless surface but later corrected it. The correct formula for acceleration is derived, leading to an answer of approximately 0.70 m/s², with emphasis on proper unit usage and significant figures. The conversation highlights the importance of accurately interpreting problem statements in physics.
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Homework Statement


An object with a mass of 10 kg (this mass is m1) rests on a horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass of 4.0 kg (this mass is m2). The coefficient
of static friction between m1 and the horizontal surface
is 0.50, and the coefficient of kinetic friction is 0.30. If the system is set in motion with m2 moving
downward, what will be the acceleration of the system?

Homework Equations


m1 = mass one
m2 = mass two
T = tension
μ = coefficient of friction
g = gravity
a = acceleration
R = normal reaction

I got this helpful hint at another site, but I still can't figure it out...

For m2:
m2 g - T = m2 a ...(1)

Resolving horizontally and vertically for m1:
T - μR = m1 a ...(2)
R = m1 g ...(3)

Substitute for R from (3) in (2):
T - μm1 g = m1 a ...(4)

Add (1) & (4):
(m2 - μm1)g = (m1 + m2)a
a = (m2 - μm1)g / (m1 + m2)
= (4 - 0.3 * 10)g / (10 + 4)
= g / 14
= 0.701 m/s.



The Attempt at a Solution



And this is where I run into the problem. I'm not sure what to do, even with the helpful hint provided on another site.
 
Last edited:
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You initially stated that m1 rests on a frictionless surface, then you give coefficients of static and kinetic friction for m1 and the surface. How is this so?
 
Oops, that was my mistake. I've edited the problem - it isn't frictionless. I'm very sorry about that...
 
Don't sweat it, I just wasn't sure what the problem was trying to accomplish. I'm not sure what you're having trouble on, you've already arrived at the correct answer. The only problem is your units and sig figs. It should be .70 m/s^2.
 
Oh, okay. Thanks a for the assistance.
 
\ W_2 - F_f = (m_1+ m_2) \cdot a eq (1)

\ N = m_1 \cdot g eq (2)

\ F_f = \mu \cdot N eq (3)

\ F_f = \mu \cdot m_1 \cdot g eq (4) from (2) into (3)

\ W_2 = m_2 \cdot g eq (5)

\ m_2 \cdot g - \mu \cdot m_1 \cdot g = (m_1+ m_2) \cdot a

\frac{\((m2 - \mu \cdot m1)} {\((m2 +m1)} g = 0.701
 
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