A wooden block A of mass 4.0 kg slides on a frictionless table when pulled using a massless string and pulley array by a hanging box B of mass 5.0 kg,as shown in the figure.What is the acceleration of block A as it slides on the frictionless table? Hint: Think carefully about the acceleration constraint. A)4.1 m/s² B)3.5 m/s² C)3.1 m/s² D)2.7 m/s²

Question

Quizplus · Accepted Answer

Since the pulley system is massless and frictionless, the tension in the string connecting the two boxes is the same throughout. Let's call this tension T. 
Now, let's consider the forces acting on block A. The only force acting on it is the tension T, which is pulling it to the right. Therefore, the net force on A is T. 
Using Newton's second law, the acceleration of block A is given by a = F_net/mass = T/mass_A. 
Next, let's consider the forces acting on box B. There are two forces acting on box B: its weight (m_B*g) pulling it down, and the tension T pulling it up. Therefore, the net force on B is F_net = T - m_B*g. 
Using Newton's second law again, the acceleration of box B is given by a = F_net/mass = (T - m_B*g)/m_B. 
Since the two boxes are connected by the same string, they must have the same acceleration (think about what would happen if one box was accelerating faster than the other). Therefore, we can equate the two expressions for acceleration: 
T/mass_A = (T - m_B*g)/m_B 
Solving for T, we get T = m_A/(m_A + m_B)*g. Substituting this into the expression for acceleration of block A, we get: 
a = T/mass_A = (m_A/(m_A + m_B)*g)/m_A = g/(m_A + m_B)*m_A 
Plugging in the numbers, we get: 
a = 9.8/(4 + 5)*4 = 4.1 m/s^2. 
Therefore, the correct answer is A.

A Wooden Block a of Mass 4