Transferring Artwork To Uneven Wood Surfaces

A block of mass 3kg is placed on a rough surface. The coefficient of static friction between two surfaces is 0.2 then what is the minimum horizontal force required to move the block?

As stated, there is not sufficient information to answer.What is the acceleration due to gravity whereas the experiment is performed?If you can assume 1 Earth standard gravity, the the block is pressed down onto the surface by a force of 9.8 * 3 N. The coefficient of friction presents a retarding force of 0.2 * 29.4, or 5.88N

If two blocks having mass (m1) and (m2) are pushed with a force F on a frictionless surface and begin moving with a uniform acceleration 'a', what would be the force between the two blocks?

The question can be solved by simple application of Newton's Second Lawof Motion.Using Newton's Second Law,it is established that[math]F_{ext}=m_{com}\,a[/math],where [math]F_{ext}[/math] is net external force acting on the system, [math] a[/math] is the acceleration produced and [math]m_{com}[/math] is mass (of center of mass of) the system(which is basically total mass of the system).Now consider both [math]m_{1}[/math] and [math]m_2[/math] as a system, so net external force acting on this system is [math]F[/math] and[math]m_{com}=m_1+m_2[/math].[math]\Rightarrow F=(m_1+m_2)a[/math][math]\Rightarrow a=\dfrac{F}{m_1+m_2}[/math].Now to find normal force between two blocks we have to consider individual blocks as a system because if you consider both of them as a single system, the normal force becomes an internal force and Newtons's Law just talks about external force.Also note that from Newton's Third Law, the normal force will act on both blocks and will be of same magnitude but in opposite direction.So lets assume that a normal force [math]F_n[/math] acts on the system.Clearly this normal force will drive [math]m_2[/math] but will oppose motion of [math]m_1[/math] (Why?).So if you consider [math]m_2[/math] as your system, net external force is [math]F_N[/math] ,mass of (center of mass of) the system is [math]m_2[/math] and acceleration is [math]a[/math].[math]\Rightarrow F_n=m_2a[/math][math]\Rightarrow \boxed{F_n=\dfrac{m_2F}{m_1+m_2}} [/math].And if you consider [math]m_1[/math] as your system,net external force is [math]F-F_N[/math] , mass of( center of mass of) the system is [math]m_1[/math] and acceleration is [math]a[/math].[math]\Rightarrow F-F_n=m_1a[/math][math]\Rightarrow F_n=F-m_1a[/math][math]\Rightarrow \boxed{F_N=\dfrac{m_2F}{m_1+m_2}}[/math]Hence required answer is [math]\boxed{\dfrac{m_2F}{m_1+m_2}}[/math].Few questions for you:-Is the given condition [math]m_1>m_2[/math] relevant? Or the given question can be solved if this condition is not given.Why is acceleration of both the masses combined is same as acceleration of individual mass?