Question

problem 6: two masses are attached by a string. one of the masses slides on a frictionless table while the other one hangs from a pulley. such a system is shown in the figure below. the masses are (m_1 = 320 g) and (m_2 = 280 g). find the acceleration of each mass ((a =?)) and the tension on the string ((t =?)). again, do not forget to draw the free - body diagrams as well as the necessary equations (from newtons second law) describing the two masses.

Explanation:

Step1: Analyze forces on $M_1$

For mass $M_1$ on the friction - less table, the only horizontal force is the tension $T$. According to Newton's second law $F = Ma$, we have $T = M_1a$.

Step2: Analyze forces on $M_2$

For mass $M_2$ hanging from the pulley, the net force acting on it is $M_2g - T$, where $g = 9.8\ m/s^{2}$. Using Newton's second law $F = Ma$, we get $M_2g - T = M_2a$.

Step3: Solve the system of equations

Substitute $T = M_1a$ into $M_2g - T = M_2a$. We have $M_2g - M_1a = M_2a$. Rearranging for $a$ gives $M_2g=(M_1 + M_2)a$. So, $a=\frac{M_2g}{M_1 + M_2}$.
First, convert masses to kg: $M_1=0.32\ kg$ and $M_2 = 0.28\ kg$. Then $a=\frac{0.28\times9.8}{0.32 + 0.28}=\frac{2.744}{0.6}\approx4.57\ m/s^{2}$.

Step4: Find the tension $T$

Substitute $a$ into $T = M_1a$. So $T = 0.32\times4.57\approx1.46\ N$.

Answer:

The acceleration $a\approx4.57\ m/s^{2}$ and the tension $T\approx1.46\ N$.