Given,
Mass of box $(m) = 6\;kg$
(i) Speed of box $(u) = 0.35\;m/s$
Coefficient of kinetic friction $(\mu_k) = 0.12$
Frictional Force $(F) = \;?$
We have,
$\mu_k = $$\frac{F_k}{R}$ ⇒ $F_k = \mu_k\;R = \mu_k\;m.g$
Now,
$F_u = F + F_k$ [∵ for Uniform Motion, $a = 0$]
$F_u = m.a + F_k = \mu_k\;m.g = 0.12\;*\; 6\;*\;10 = 7.2\;N$
(ii) When the box moves with acceleration $0.18\;m/s^2$. Then
$F_u = m.a + m_k\;m/g = 6\;*\;0.18 + 0.12\;*\;6\;*10 = 8.28\;N$
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