Question

A cord passing over a frictionless, massless pulley has a 4 kg. object tied to one end and 12 kg object tied to the other. compute the acceleration and the tension in the cord.

Answer 1

(a) Let's apply the Newton's Second Law of Motion:

$$T-m_1g=m_1a,$$ $$T-m_2g=-m_2a.$$

Eliminating $T$, we get:

$$(m_2-m_1)g=(m_1+m_2)a,$$ $$a=\dfrac{(m_2-m_1)g}{m_1+m_2},$$ $$a=\dfrac{(12\ kg-4\ kg)\cdot9.8\ \dfrac{m}{s^2}}{4\ kg+12\ kg}=4.9 \ \dfrac{m}{s^2}.$$

(b) We can find the tension in the cord by substituting $a$ into the first equation:

$$T=m_1(a+g)=4\ kg\cdot(4.9 \ \dfrac{m}{s^2}+9.8 \ \dfrac{m}{s^2})=58.8\ N.$$