Suppose a hanging 1.0 kg lab mass is attached to a 4.0 kg block on the table.
a. If the coefficient of kinetic friction, uk is 0.20, what is the acceleration of the block?
b. What would be the minimum value of the coefficient of static friction, Us, in order for the block to remain motionless?
a. The acceleration of the block is 0.25 m/s^2.
b. The minimum value of the coefficient of static friction, Us, in order for the block to remain motionless is 0.50.
To solve these problems, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).
a. To find the acceleration of the block, we first need to calculate the net force acting on it. The only horizontal forces acting on the block are the force of kinetic friction and the tension in the string.
The force of kinetic friction (Fk) can be calculated using the equation Fk = uk * N, where uk is the coefficient of kinetic friction and N is the normal force.
The normal force (N) is equal to the weight of the block, which is given by N = mg, where m is the mass of the block and g is the acceleration due to gravity.
In this case, the mass of the block is 4.0 kg, so the weight is N = 4.0 kg * 9.8 m/s^2 = 39.2 N.
Therefore, the force of kinetic friction is Fk = 0.20 * 39.2 N = 7.84 N.
Since the tension in the string is equal to the weight of the hanging mass (10 N), the net force can be calculated as Fnet = T - Fk = 10 N - 7.84 N = 2.16 N.
Now, we can find the acceleration by dividing the net force by the mass of the block: a = Fnet / m = 2.16 N / 4.0 kg = 0.54 m/s^2.
So, the acceleration of the block is 0.54 m/s^2.
b. To calculate the minimum value of the coefficient of static friction, we need to consider the maximum force of static friction (Fs). The maximum force of static friction can be calculated using the equation Fs = Us * N, where Us is the coefficient of static friction.
For the block to remain motionless, the force of static friction must be equal and opposite to the applied force, which in this case is the force due to the hanging mass (10 N).
So, we have Fs = 10 N.
Substituting the value of the weight N = 4.0 kg * 9.8 m/s^2 = 39.2 N, we get Us * 39.2 N = 10 N.
Therefore, the minimum value of the coefficient of static friction (Us) is Us = 10 N / 39.2 N = 0.255.
So, the minimum value of the coefficient of static friction required for the block to remain motionless is 0.255.
To find the acceleration of the block, we need to consider the forces acting on it.
a. The force pulling the block is the tension in the string, which is equal to the weight of the hanging mass. The weight is given by:
Weight = mass x acceleration due to gravity
Weight = 1.0 kg x 9.8 m/s^2
Weight = 9.8 N
The force of kinetic friction acting on the block can be calculated as:
Force of kinetic friction = coefficient of kinetic friction x normal force
The normal force is equal to the weight of the block, which is:
Normal force = mass x acceleration due to gravity
Normal force = 4.0 kg x 9.8 m/s^2
Normal force = 39.2 N
Therefore, the force of kinetic friction is:
Force of kinetic friction = 0.20 x 39.2 N
Force of kinetic friction = 7.84 N
Now, we can determine the net force acting on the block:
Net force = tension - force of kinetic friction
Net force = 9.8 N - 7.84 N
Net force = 1.96 N
Finally, we can calculate the acceleration using Newton's second law, which states that the net force is equal to the mass multiplied by the acceleration:
Net force = mass x acceleration
1.96 N = 4.0 kg x acceleration
Solving for acceleration:
acceleration = 1.96 N / 4.0 kg
acceleration = 0.49 m/s^2
Therefore, the acceleration of the block is 0.49 m/s^2.
b. To find the minimum value of the coefficient of static friction for the block to remain motionless, we need to consider the maximum force of static friction.
The maximum force of static friction can be calculated as:
Force of static friction = coefficient of static friction x normal force
To keep the block motionless, the force of static friction must be equal to the force pulling the block (tension). Since the force pulling the block is the weight of the hanging mass, the force of static friction must be equal to 9.8 N.
Therefore, the equation becomes:
9.8 N = coefficient of static friction x 39.2 N
Solving for the coefficient of static friction:
coefficient of static friction = 9.8 N / 39.2 N
coefficient of static friction = 0.25
Therefore, the minimum value of the coefficient of static friction for the block to remain motionless is 0.25.