Question

A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disk of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.21 m off the ground. What speed does this block have when it hits the ground?

Answer 1

The answer is "0.2711 m/s".

Step-by-step explanation:

Potential energy = Kinetic energy + Potential energy

`m_1 gh =(1)/(2) m_1v^2 +(1)/(2) m_2v^2 + (1)/(2) I\omega^2 + m_1gh\\\\`

`(m_1- m_2)gh =(1)/(2) m_1v^2 +(1)/(2) m_2v^2 +(1)/(2) I\omega^2\\\\2(m_1 - m_2)gh = m_1v^2 + m_1v^2 + I\omega^2\\\\solid \ disk (I) = (1)/(2) \ \ M r^2 \\\\`

When there is no slipping, \omega =\frac{ v]{r}\\\\

`2(m_1 - m_2)gh = m_1v^2 + m_2v^2 + ((1)/(2) Mr^2) ((v)/(r))^2\\\\2(m_1 -m_2)gh = m_1v^2 + m_2v^2 + (1)/(2) Mv^2\\\\4(m_1 -m_2)gh = 2m_1v^2 + 2m_2v^2 + Mv^2\\\\4(m_1 - m_2)gh = (2m_1 + 2m_2 + M) v^2\\\\`

`v^2 = (4(m_1 - m_2)gh)/((2m_1 + 2m_2 + M))v`

`v^2 = (4 (0.25 \ kg - 0.20 \ kg) (9.8 (m)/(s^2)) (0.21 m))/( (2 * 0.25 kg + 2 * 0.20 kg + 0.50 kg))`

`=(0.1029)/(1.4) \ \ (m^2)/(s^2)\\\\=0.0735\ \ (m^2)/(s^2)\\\\= 0.2711 \ (m)/(s)`