Question

1. in figure 6, a cord of negligible mass connects a 0.500 kg mass to a 1.00 kg mass. there is negligible friction in the system. (a) at what rate will the masses accelerate? (b) what is the tension in the cord while the masses are accelerating?

Explanation:

Step1: Apply Newton's second - law to the system

Let the acceleration of the system be $a$. The net force acting on the system of two masses $m_1 = 0.500\ kg$ and $m_2=1.00\ kg$ is $F_{net}=(m_2 - m_1)g$ (taking the downward - direction for the heavier mass as positive). According to Newton's second law $F = ma$, where $m=m_1 + m_2$. So, $(m_2 - m_1)g=(m_1 + m_2)a$.

Step2: Solve for the acceleration $a$

We can re - arrange the equation $(m_2 - m_1)g=(m_1 + m_2)a$ to find $a=\frac{m_2 - m_1}{m_1 + m_2}g$. Substituting $m_1 = 0.500\ kg$, $m_2 = 1.00\ kg$ and $g = 9.8\ m/s^2$, we get $a=\frac{1.00 - 0.500}{1.00+0.500}\times9.8\ m/s^2=\frac{0.500}{1.500}\times9.8\ m/s^2\approx3.27\ m/s^2$.

Step3: Apply Newton's second - law to one of the masses to find the tension

Let's consider the $0.500\ kg$ mass. The forces acting on it are the tension $T$ upwards and its weight $m_1g$ downwards. According to Newton's second law $F = ma$, so $T - m_1g=m_1a$. Then $T=m_1(g + a)$. Substituting $m_1 = 0.500\ kg$, $g = 9.8\ m/s^2$ and $a\approx3.27\ m/s^2$, we get $T=0.500\times(9.8 + 3.27)\ N=0.500\times13.07\ N = 6.535\ N\approx6.54\ N$.

Answer:

(a) $a\approx3.27\ m/s^2$
(b) $T\approx6.54\ N$