Question

HELP ME PLEASE I BET

Two blocks are attached to a light string. Block À (m A - 9 kg) is at rest on a frictionless table. Block B (mg - 1.24 kg) hangs off the edge of the table, as shown. Both blocks are released from rest
How much tension is in the string?

Answer 1

Approximately
`10.7\; {\rm N}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Let the magnitude of this tension be
`F_{\text{tension}}`.

In block
`\text{A}`, this tension points towards the edge of the table and is unbalanced. Hence, the net force on block
`\text{A}\!` would be equal to the tension,
`F_{\text{tension}}`. The acceleration of block
`\text{A}\!\!` would be
`(F_\text{tension}) / (m_{\text{A}})`.

In block
`\text{B}`, this tension points upwards and acts against the weight
`m_{\text{B}}\, g` of this block. Hence, the net force on block
`\text{B}\!` would be
`(m\, g - F_{\text{tension}})`. The acceleration of block
`\text{B}\!\!` would be
`(m\, g - F_{\text{tension}}) / (m_{\text{B}})`.

If the light string between the two blocks is inelastic, the acceleration of the two blocks should be equal in magnitude. Hence:

`\displaystyle \frac{F_\text{tension}}{m_{\text{A}}} = \frac{m_{\text{B}}\, g - F_{\text{tension}}}{m_{\text{B}}}`.

Rearrange this equation and solve for
`F_{\text{tension}}`:

`\displaystyle \frac{F_\text{tension}}{m_{\text{A}}} = g - \frac{F_{\text{tension}}}{m_{\text{B}}}`.

`\begin{aligned} F_{\text{tension}} &= \frac{g}{(1/m_{\text{A}}) + (1 / m_{\text{B}})} \\ &= \frac{9.81\; {\rm m\cdot s^(-2)}}{(1 / (9\; {\rm kg})) + (1 / (1.24\; {\rm kg}))} \\ &\approx 10.7\; {\rm N}\end{aligned}`.