A body of mass m_2 = 4 kg lies on a wheeled platform of mass…
A body of mass $m_2 = 4\ \mathrm{kg}$ lies on a wheeled platform of mass $m_1 = 40\ \mathrm{kg}$; the coefficient of friction between them is $\mu = 0.2$. A horizontal force acts on the body. With $F_1 = 2\ \mathrm{N}$ the body does not move relative to the platform; with $F_2 = 100\ \mathrm{N}$ it slides. Find the friction force and the platform's acceleration in both cases. The platform rolls without friction.
Explanation of Method
A body on a wheeled platform.
Step 1: Case 1: $F_1 = 2\ \mathrm{N}$, moving together
\[ a_1 = \frac{F_1}{m_1 + m_2} = \frac{2}{44} \approx 0.045\ \mathrm{m/s^2}. \]
Friction is the only horizontal force on the platform, so it is the force that gives the platform this acceleration: $f_1 = m_1 a_1 = 40 \cdot 0.0455$.
\[f_1 \approx 1.82\ \mathrm{N}, \qquad a_1 \approx 0.045\ \mathrm{m/s^2}\]
That is below the maximum $\mu m_2 g = 7.85\ \mathrm{N}$, so the body indeed does not slip.
Step 2: Case 2: $F_2 = 100\ \mathrm{N}$, sliding
Now friction has its sliding value, and it alone drives the platform: $f_2 = \mu m_2 g = 0.2 \cdot 4 \cdot 9.81 \approx 7.85\ \mathrm{N}$.
\[ a_2 = \frac{f_2}{m_1}. \]
\[f_2 \approx 7.85\ \mathrm{N}, \qquad a_2 \approx 0.196\ \mathrm{m/s^2}\]
(The body itself accelerates at $(100 - 7.85)/4 \approx 23\ \mathrm{m/s^2}$.)
When does slipping start?
Moving together, the platform needs $f = m_1 F/(m_1 + m_2)$. This reaches $\mu m_2 g$ at $F = \mu m_2 g\,(m_1 + m_2)/m_1 \approx 8.6\ \mathrm{N}$.
Result
\[\boxed{\,f_1 \approx 1.82\ \mathrm{N}, \qquad a_1 \approx 0.045\ \mathrm{m/s^2}\,}\]
\[\boxed{\,f_2 \approx 7.85\ \mathrm{N}, \qquad a_2 \approx 0.196\ \mathrm{m/s^2}\,}\]
Remark
About the original solution. In case 1 it set the friction equal to the applied force, 2 N. That would leave zero net force on the body, yet it accelerates. The friction is only the part of the force passed on to the platform, $m_1 a_1 \approx 1.82\ \mathrm{N}$. Case 2 is correct.