A 5.00-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 0.650. What is the maximum horizontal force that can be applied before the 5.00-kg block begins to slip relative to the 12.0-kg block, if the force is applied to the smaller block?

To find the maximum horizontal force that can be applied before the 5.00-kg block begins to slip relative to the 12.0-kg block, we need to consider the concept of static friction.

Static friction is the force that opposes the motion or potential motion of an object at rest. The maximum static friction force can be calculated using the equation:

Fstatic = μs * N

where Fstatic is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force.

In this case, the normal force acting on the 5.00-kg block is equal to its weight, since the block is on a frictionless table and there are no vertical forces acting on it. The weight of an object can be calculated using the equation:

Weight = mass * acceleration due to gravity

Thus, the weight of the 5.00-kg block is given by:

Weight of 5.00-kg block = 5.00 kg * 9.8 m/s^2

Next, we can calculate the maximum static friction force between the two blocks using the coefficient of static friction and the normal force.

Fstatic = 0.650 * (Weight of 5.00-kg block + Weight of 12.0-kg block)

Finally, the maximum horizontal force that can be applied before the 5.00-kg block begins to slip is equal to the maximum static friction force between the blocks.

Therefore, the maximum horizontal force is Fstatic as calculated above.