TWO BOXES A AND B ARE LYING ON THE TABLE AND ARE CONNECTED BY A PIECE OF STRING. THE MASS OF BOX A IS 3KG AND B IS 2KG. A 10N PULLING FORCE, POINTING TO THE RIGHT IS APPLIED TO BOX X CAUSING THE TWO BOXES TO MOVE TOGETHER AT CONSTA SPEED. a) determine the net force on A & B b) calculate the coefficient of friction between the table and boxes. c) calculate the friction force exerted on box A and B.
Box X ?
To answer these questions, we need to understand the concept of Newton's laws of motion and the forces acting on the objects.
a) Determine the net force on A and B:
When two objects are connected and moving together at a constant speed, we can conclude that the net force acting on them is zero. This is based on Newton's first law of motion, which states that an object will remain at rest or in constant motion unless acted upon by an external force. Since the two boxes are moving together at a constant speed, the net force on both A and B is zero.
b) Calculate the coefficient of friction between the table and boxes:
In this case, the only force acting on the boxes is the pulling force of 10N. However, to maintain the constant speed, there must be a frictional force acting in the opposite direction. This frictional force can be determined using the equation:
Frictional force (F_friction) = μ * Normal force
Here, μ represents the coefficient of friction, and the normal force is the force exerted by the table on the boxes perpendicular to the surface.
Since the boxes are not accelerating vertically, the normal force exerted by the table on the boxes is equal to the gravitational force acting on them.
Normal force (F_normal) = mass * g
For box A: F_normal A = 3kg * 9.8 m/s^2 = 29.4 N
For box B: F_normal B = 2kg * 9.8 m/s^2 = 19.6 N
Next, we can calculate the coefficient of friction:
F_friction = μ * F_normal
10N (pulling force) = μ * (29.4N + 19.6N)
10N = μ * 49N
μ = 10N / 49N
μ = 0.2041
Therefore, the coefficient of friction between the table and the boxes is approximately 0.2041.
c) Calculate the friction force exerted on box A and B:
Since the net force acting on both boxes is zero, the total frictional force acting on the boxes will be 10N (in the opposite direction of the pulling force).
Since the masses of boxes A and B are different (3kg and 2kg respectively), the friction forces acting on them will also be different. We can calculate the friction force using the equation:
Frictional force (F_friction) = μ * F_normal
For box A: F_friction A = μ * F_normal A = 0.2041 * 29.4N = 6.001N
For box B: F_friction B = μ * F_normal B = 0.2041 * 19.6N = 4.001N
Therefore, the friction force exerted on box A is approximately 6.001N, while the friction force exerted on box B is approximately 4.001N.