Question

Two blocks, one of mass 5 kg and the other of mass 2 kg, are attached to opposite ends of a light string and hung vertically from a massless, frictionless pulley. Initially the heavier block is held in place a distance 2.5 m above the floor, the lighter block is just touching the floor, and the cord is taut. Then the heavier block is released and comes crashing to the floor while the cord slackens and the lighter block continues to rise. What is the maximum height reached by the lighter block

Answer 1

`H_(max)=3.4m`

Step-by-step explanation:

From the question we are told that:

Mass 1
`m_1=5kg`

Mass
`m_2=2kg`

Distance above floor
`d=2.5m`

Generally the equation for Conservation of energy is mathematically given by

`0.5m_1v^2+0.5m_2v^2=2mg`

`0.5m_1v^2+0.5m_2v^2=2g(m_1-m_2)`

`v^2(0.5*m_1+0.5*m_2)=2*g(m_1-m_2)`

`v^2(0.5*5+0.5*2) = 2 * 9.8 * (5 - 2)`

`v^2=(58.8)/(3.5)`

`v=4.1m/s`

Generally the equation for The maximum height of lighter block is mathematically given by

`H_(max)=d+(v^2)/(2g)`

`H_(max)=2.5+(16.2)/(2*9.81)`

`H_(max)=3.4m`