Kinetic friction force

Review the object with a mass of 3 kg :

∑F = m.a

T-W = m.a

T = m.a + m.g ... (1)

Now, review the object with a mass of 2 kg :

∑F = m.a

F - T - f = m.a

F - (m.a + m.g) - μ.N = m.a

F = m.g + μ.N

F = 40 N

Hello, in this system of 2 masses involved, as uniform motion means = v = constant, a = 0 m/s/s,

for the 2kg mass, by applying Fnett = ma,

F - ( T + fk ) = 2(0)

F - T - (0.5 x 2 x 10 ) = 0

F = T + 10

for the 3kg mass,

T - W = ma

T = 3(0) + (3)(10) = 30 N

As T = 30N,

F = 30 + 10 = 40N,

Thanks....

Since both the blocks are in uniform motion,

$F_{net,3kg block}=0$

$T_{rope}-W_{3kg block}=0$

$T_{rope}=(3kg)(10N/kg)=30N$

On the other hand,

$F_{net,2kg block}=0$

$F-(T_{rope}+f_k)=0$

$F = 30N+μ_kF_N$

$F = 30N+(0.5)(20N)=40N$