Two blocks (on top of one another), each of mass m = 2.4 kg, are pushed along the horizontal surface of a table by a horizontal force of magnitude 6.8 N, directed to the right. The blocks move together to the right at constant velocity.
(a) Find the frictional force exerted on the lower block by the table.
(b) Find the coefficient of kinetic friction between the surface of the block and the table.
(c) Find the frictional force acting on the upper block.
To solve this problem, we can use Newton's second law of motion, which states that the net force on an object is equal to the product of its mass and acceleration. Since the blocks are moving together at a constant velocity, we know that the net force acting on both blocks is zero.
(a) To find the frictional force exerted on the lower block by the table, we can use the equation F_friction = μ * F_normal, where F_friction is the frictional force, μ is the coefficient of friction, and F_normal is the normal force.
Since the blocks are at rest vertically, the normal force acting on each block is equal to the gravitational force pulling them downwards, which is given by F_gravity = m * g, where m is the mass of each block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, the normal force and the frictional force are equal: F_normal = F_friction.
Substituting the given mass m = 2.4 kg, we have:
F_gravity = m * g
F_gravity = 2.4 kg * 9.8 m/s^2
F_gravity ≈ 23.52 N
So, the frictional force exerted on the lower block by the table is approximately 23.52 N.
(b) To find the coefficient of kinetic friction between the surface of the block and the table, we can use the equation F_friction = μ * F_normal and substitute the known values:
F_friction = μ * F_normal
23.52 N = μ * 23.52 N
Simplifying the equation, we find:
μ = F_friction / F_normal
μ = 23.52 N / 23.52 N
μ ≈ 1
Therefore, the coefficient of kinetic friction between the surface of the block and the table is approximately 1.
(c) Since the blocks are moving together with constant velocity, the frictional force acting on the upper block is equal to the frictional force acting on the lower block. So, the frictional force acting on the upper block is also approximately 23.52 N.
To find the answers to these questions, we need to analyze the forces acting on the blocks. There are three forces at play here: the pushing force, the gravitational force, and the frictional force.
(a) To find the frictional force exerted on the lower block by the table, we need to first determine the net force acting on the blocks. Since the blocks are moving at a constant velocity, we know that the net force is zero. This means that the pushing force is balanced by the frictional force and the gravitational force.
The gravitational force acting on each block can be found using the formula F_gravity = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
For each block, the gravitational force is F_gravity = (2.4 kg)(9.8 m/s^2) = 23.52 N.
Since the blocks are moving at a constant velocity, the pushing force must be equal in magnitude but opposite in direction to the frictional force. Therefore, the frictional force exerted on the lower block by the table is 6.8 N to the left.
(b) To find the coefficient of kinetic friction between the surface of the block and the table, we can use the formula f_friction = μ_k * F_normal, where μ_k is the coefficient of kinetic friction and F_normal is the normal force exerted by the table on the block.
The normal force exerted on the lower block by the table is equal to the gravitational force acting on the lower block, which is 23.52 N.
Therefore, we can rearrange the formula to solve for μ_k: μ_k = f_friction / F_normal.
Using the values we have, μ_k = 6.8 N / 23.52 N ≈ 0.289.
Therefore, the coefficient of kinetic friction between the surface of the block and the table is approximately 0.289.
(c) To find the frictional force acting on the upper block, we know that the pushing force is balanced by the frictional force and the gravitational force for the entire system (both blocks together). Therefore, the frictional force acting on the upper block must be equal in magnitude but opposite in direction to the pushing force.
Since the pushing force is 6.8 N to the right, the frictional force acting on the upper block is 6.8 N to the left.