6. Two blocks of masses and m_{2} with m_{1} > m_{2} are placed in contact with each other on a m_{1} frictionless, horizontal surface. A constant horizontal force F is applied to m_{1}
To find the acceleration of the system, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
In this scenario, the net force acting on the system is the force applied to m₁ (F), and the combined mass of the two blocks is (m₁ + m₂). The force of m₂ on m₁ will be equal to the force of m₁ on m₂ due to Newton's third law of motion.
Therefore, the net force acting on the system is:
F_net = F - F₂ = F - F₁
where F₁ is the force exerted by m₂ on m₁ and F₂ is the force exerted by m₁ on m₂.
Since the surface is frictionless, the force of m₁ on m₂ is equal to the force of m₂ on m₁, which means F₁ = F₂.
Therefore:
F_net = F - F₁ - F₁ = F - 2F₁
Now, we need to find F₁. We know that the force of friction (F_friction) between two objects in contact is equal to the coefficient of friction (μ) multiplied by the normal force (F_normal). In this case, the normal force on m₁ is m₁ * g (where g is the acceleration due to gravity). As the surface is frictionless, the normal force will be equal to the weight of m₁, which is m₁ * g.
So, F_friction = μ * (m₁ * g)
Since F_friction is equal to F₁, we can write:
F₁ = μ * (m₁ * g)
Now, substituting F₁ in the equation for F_net:
F_net = F - 2 * μ * (m₁ * g)
Finally, using Newton's second law of motion:
F_net = (m₁ + m₂) * a
where 'a' is the acceleration of the system.
Now, we have two equations:
F - 2 * μ * (m₁ * g) = (m₁ + m₂) * a
From these equations, you can solve for the acceleration of the system.