Question

1. two blocks are connected by a string and lie on a frictionless tabletop with m₁ = 1.4 kg. the tension force between blocks is t = 24 n, while the force applied is f = 52 n.

a. draw a free - body diagram for the system and each block showing all applied forces to scale. next to each diagram show the direction of the acceleration of that object.
b. find the acceleration of each object.
c. find the mass of m₂.

Explanation:

Step1: Find acceleration via $m_1$

For $m_1$, net force is tension $T$. Use $F=ma$:
$$a = \frac{T}{m_1} = \frac{24}{1.4} \approx 17.14 \, \text{m/s}^2$$

Step2: Find total mass of system

For the combined system, net force is $F_{app}$. Use $F=ma$:
$$M_{total} = \frac{F_{app}}{a} = \frac{52}{\frac{24}{1.4}} = \frac{52 \times 1.4}{24} \approx 3.03 \, \text{kg}$$

Step3: Calculate $m_2$

Subtract $m_1$ from total mass:
$$m_2 = M_{total} - m_1 = \frac{52 \times 1.4}{24} - 1.4$$

Answer:

Part a (Free-Body Diagrams):

• System FBD: A single combined block with upward normal force $F_N$ equal to total weight $(m_1+m_2)g$ downward, and rightward applied force $F_{app}=52\,\text{N}$. Acceleration direction is rightward.
• $m_1$ FBD: Block with upward normal force $F_{N1}=m_1g$ downward, rightward tension force $T=24\,\text{N}$. Acceleration direction is rightward.
• $m_2$ FBD: Block with upward normal force $F_{N2}=m_2g$ downward, rightward applied force $F_{app}=52\,\text{N}$, leftward tension force $T=24\,\text{N}$. Acceleration direction is rightward.

Part b (Acceleration):

$a \approx 17.14 \, \text{m/s}^2$ (same for both blocks)

Part c (Mass of $m_2$):

$m_2 \approx 1.63 \, \text{kg}$