AP Physics 1 Exam Question:

A 2 kg mass is placed on a table and is connected to a pulley by a string. The mass falls vertically downward from a height of 1 meter. The string passes over the pulley and is connected to a hanging 1 kg mass on the other side. The coefficient of kinetic friction between the table and the 2 kg mass is 0.2. Calculate the final velocity of the 2 kg mass just before it reaches the table. Assume g = 9.8 m/s².

(A) 1.22 m/s (B) 2.55 m/s (C) 3.89 m/s (D) 5.1 m/s

Solution:

First, let's analyze the forces acting on the system. Since the 1 kg mass is lighter than the 2 kg mass, it will cause the 2 kg mass to accelerate downward.

The forces acting on the 1 kg mass are:

1. Weight (mg): The mass of the 1 kg mass multiplied by acceleration due to gravity (9.8 m/s²), acting downward.
2. Tension in the string (T): The tension in the string pulling the 1 kg mass upward.

The forces acting on the 2 kg mass are:

1. Weight (2mg): The mass of the 2 kg mass multiplied by acceleration due to gravity (9.8 m/s²), acting downward.
2. Normal force (N): The force exerted by the table, acting upward.
3. Frictional force (f): The force of kinetic friction between the table and the 2 kg mass, acting opposite to the direction of motion.
4. Tension in the string (T): The tension in the string pulling the 2 kg mass upward.

Since the 1 kg mass is being pulled downward, it will cause the 2 kg mass to accelerate downward. Therefore, we have the following relationship between their accelerations:

a_1kg = a_2kg

Now, let's apply Newton's second law to both masses separately.

For the 1 kg mass:

m_1kg * a_1kg = T - m_1kg * g ...(1)

For the 2 kg mass:

m_2kg * a_2kg = T - m_2kg * g - f ...(2)

Next, let's calculate the frictional force (f), using the coefficient of kinetic friction and the normal force.

f = μ_k * N

Since the 2 kg mass is in contact with the table, the normal force (N) acting on it is equal to its weight.

N = m_2kg * g

Substituting this into the equation for frictional force, we have:

f = μ_k * (m_2kg * g)

Substituting the value of the coefficient of kinetic friction (μ_k = 0.2), we have:

f = 0.2 * (2 kg * 9.8 m/s²)

f = 3.92 N

Now, let's substitute this value into equation (2) and solve for acceleration:

m_2kg * a_2kg = T - m_2kg * g - f

2 kg * a_2kg = T - 2 kg * 9.8 m/s² - 3.92 N

Simplifying, we get:

2 kg * a_2kg = T - 19.6 N - 3.92 N

2 kg * a_2kg = T - 23.52 N

Next, we can substitute the value of the tension (T) from equation (1) into equation (2):

m_1kg * a_1kg = T - m_1kg * g

1 kg * a_1kg = T - 1 kg * 9.8 m/s²

1 kg * a_1kg = T - 9.8 N

Rearranging the equation, we get:

T = 1 kg * a_1kg + 9.8 N

Substituting this value of tension (T) into equation (2), we have:

2 kg * a_2kg = (1 kg * a_1kg + 9.8 N) - 23.52 N

2 kg * a_2kg = 1 kg * a_1kg - 13.72 N

Since we know the relationship between the accelerations (a_1kg = a_2kg), we can simplify the equation:

2 kg * a_2kg = a_2kg - 13.72 N

Rearranging, we get:

a_2kg = - 13.72 N / 1 kg

a_2kg = - 13.72 m/s²

The negative sign indicates that the 2 kg mass is accelerating downward.

Finally, we can find the final velocity (v_final) of the 2 kg mass just before it reaches the table using the kinematic equation:

v_final² = v_initial² + 2 * a * d

Considering the placement of the masses, the initial velocity is 0 m/s as the 2 kg mass is starting from rest. The distance (d) can be taken as the height from which the 2 kg mass falls (1 meter).

v_final² = 0 + 2 * (-13.72 m/s²) * 1 m

v_final² = -27.44 m²/s²

Since velocity cannot be negative in this context, we take the positive square root:

v_final = √(27.44 m²/s²)

v_final = 5.24 m/s

Therefore, the final velocity of the 2 kg mass just before it reaches the table is approximately 5.24 m/s.

Answer: (D) 5.1 m/s