Complete Question:
A 5 g block rests on a rough horizontal table. A rope is attached to the block and is pulled with a force of 11 N to the left. As a result, the block accelerates at 2 m/s^2. What is the kinetic friction between the block and the table?
Correct Answer:
The kinetic frictional force between the block and the rough surface is 10.99 N towards the right side.
[tex]F_{net}[/tex] = - 0.01 N is the resulting net force causing an acceleration of [tex]2\ m/s^2[/tex] to the left
Explanation:
According to the first law of motion by Newton, any object can be moved only when an unbalanced force acts on the object. As the block is placed on a rough horizontal table, then the applied force should be greater than the kinetic frictional force between the object and the rough surface, in order to move that block.
So, it is known that the net force acting on the block will be the difference between the force acting on the object due to the rope and the frictional force acting on the object.
[tex]F_{net} = F_{rope} -F_{friction}[/tex]
So, the force acting on the block due to rope is given as 11 N. The net force should be equal to the product of mass and acceleration of the object, as per Newton's second law of motion. As the mass of the block is 5 g and the acceleration is given as [tex]2\ m/s^2[/tex].
[tex]F_{net} = Mass * Acceleration = \frac{5}{1000} * -2 = -0.01 N[/tex]
Since, the net force and the force due to rope is acting opposite to the gravitational force ,
-0.01 = -11 - Frictional force
Frictional force = -11 + 0.01 = -10.99 N
Thus, the kinetic frictional force between the block and the rough surface is 10.99 N towards the right side.
[tex]F_{net}[/tex] = - 0.01 N is the resulting net force causing an acceleration of [tex]2\ m/s^2[/tex] to the left
