Need a Double Check / Verification

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The discussion revolves around calculating the tension in a cable and the acceleration of two connected masses, with a coefficient of friction of 0.15 between one mass and the surface. The calculations begin with the forces acting on the 2 kg and 5 kg masses, leading to two key equations derived from free body diagrams. The resulting acceleration is calculated to be 3.85 m/s², while the tension in the cable is found to be 11.9 N. However, a participant points out a mistake in the equations, indicating that the setup of the equations is incorrect. The verification of the calculations is essential for accurate results.
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In the following diagram there is a coefficient of friction, μ, of 0.15 between the 5.0 kg mass and the surface. Calculate the tension in the cable connecting the two masses and the resulting acceleration.

FBD
Physics Question 3 U1-C.gif


m1= 2 kg
m2= 5 kg
μ= .15
g= 9.8 m/s2

Equation 1 ~

Fg - FT = FN

2kg × 9.8m/s2 - 2kg(a) = FT

Equation 2 ~

Fg = FN = 5kg (9.8 m/s2)
~ μFN = .15(49) = 7.35 N

FT - FF = ma

FT = 5kg(a) + 7.35

~Set Equations Equal to Each Other~

19.6 - 2a = 5a - 7.35

26.95 = 7a

a = 3.85 m/s2

Sub into Eq to get FT on String

2 (9.8) - 2(3.85) = FT

FT = 11.9 N
 
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Inferior Mind said:
In the following diagram there is a coefficient of friction, μ, of 0.15 between the 5.0 kg mass and the surface. Calculate the tension in the cable connecting the two masses and the resulting acceleration.

FBD
View attachment 55355

m1= 2 kg
m2= 5 kg
μ= .15
g= 9.8 m/s2

Equation 1 ~

Fg - FT = FN

2kg × 9.8m/s2 - 2kg(a) = FT

Equation 2 ~

Fg = FN = 5kg (9.8 m/s2)
~ μFN = .15(49) = 7.35 N

FT - FF = ma

FT = 5kg(a) + 7.35

~Set Equations Equal to Each Other~

19.6 - 2a = 5a - 7.35

26.95 = 7a

a = 3.85 m/s2

Sub into Eq to get FT on String

2 (9.8) - 2(3.85) = FT

FT = 11.9 N

Look Good !
 
No, it isn't smart guy.

5a + 7.35 =/= 5a - 7.35
 

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