Need a Double Check / Verification

[Inferior Mind]

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

m₁= 2 kg
m₂= 5 kg
μ= .15
g= 9.8 m/s²

Equation 1 ~

F_g - F_T = F_N

2kg × 9.8m/s² - 2kg(a) = F_T

Equation 2 ~

F_g = F_N = 5kg (9.8 m/s²)
~ μF_N = .15(49) = 7.35 N

F_T - F_F = ma

F_T = 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 F_T on String

2 (9.8) - 2(3.85) = F_T

F_T = 11.9 N

 

[SammyS]

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 ~

F_g - F_T = F_N

2kg × 9.8m/s² - 2kg(a) = F_T

Equation 2 ~

F_g = F_N = 5kg (9.8 m/s²)
~ μF_N = .15(49) = 7.35 N

F_T - F_F = ma

F_T = 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 F_T on String

2 (9.8) - 2(3.85) = F_T

F_T = 11.9 N

Look Good !

 

[Inferior Mind]

No, it isn't smart guy.

5a + 7.35 =/= 5a - 7.35