Two blocks (one on top of the other), each of mass m = 2.4 kg, are pushed along the horizontal surface of a table by a horizontal force P 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.
a) friction force=mu*(2*2.4)g
b) Friction force=6.8 if constant velocity, solve for mu.
what exactly does mu stand for?
To find the frictional force exerted on the lower block by the table, we need to use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, since the blocks are moving at a constant velocity, their acceleration is zero.
(a) Since the blocks are moving at a constant velocity, the net force acting on them must also be zero. We have a horizontal force P pushing the blocks to the right, so the frictional force exerted by the table on the lower block must be equal in magnitude and opposite in direction to P.
Therefore, the frictional force exerted on the lower block by the table is 6.8 N, directed 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:
frictional force = coefficient of friction * normal force
The normal force is the force exerted by the table on the lower block, which is equal in magnitude and opposite in direction to the gravitational force acting on the lower block.
The gravitational force acting on the lower block can be found using the formula:
gravitational force = mass * acceleration due to gravity
where mass is the mass of the lower block (m = 2.4 kg) and acceleration due to gravity is approximately 9.8 m/s^2.
acceleration due to gravity = 9.8 m/s^2
gravitational force = mass * acceleration due to gravity
= 2.4 kg * 9.8 m/s^2
= 23.52 N
Since the blocks are moving together, the normal force exerted by the table on the lower block is equal in magnitude to the gravitational force, so:
normal force = 23.52 N
Now, we can plug the values of the frictional force (6.8 N) and the normal force (23.52 N) into the formula for frictional force:
frictional force = coefficient of friction * normal force
6.8 N = coefficient of friction * 23.52 N
Solving for the coefficient of friction:
coefficient of friction = 6.8 N / 23.52 N
coefficient of friction ≈ 0.29
Therefore, the coefficient of kinetic friction between the surface of the block and the table is approximately 0.29.