A 4.85-kg block is placed on top of a 12.0-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.645. What is the maximum horizontal force that can be applied before the 4.85-kg block begins to slip relative to the 12.0-kg block, if the force is applied to the more massive block?

force friction between blocks: 4.85*9.8*.645

The max acceleration: frictionforce/mass=9.8*.645 m/s^2

max horizontal force= (12+4.85)a

To find the maximum horizontal force that can be applied before the 4.85-kg block begins to slip relative to the 12.0-kg block, we need to calculate the maximum frictional force between the two blocks.

Let's start by finding the normal force acting on the 4.85-kg block. The normal force is equal to the weight of the block, which is given by the formula:

Normal force = mass × acceleration due to gravity

Plugging in the values:

Normal force = 4.85 kg × 9.8 m/s^2

Normal force = 47.63 N

Next, we can calculate the maximum static frictional force between the two blocks using the coefficient of static friction. The formula for static friction is:

Frictional force (max) = coefficient of static friction × normal force

Plugging in the values:

Frictional force (max) = 0.645 × 47.63 N

Frictional force (max) = 30.77 N

Therefore, the maximum horizontal force that can be applied before the 4.85-kg block begins to slip relative to the 12.0-kg block is 30.77 N.