Question

The block of massM inFIGURE $P7.36$ slides on a frictionless surface. Find an expression for the tension in the string

Short answer

Expression for tension is$T=\frac{Mmg}{M+m}$.

Step-by-step solution

Given information

"M" and "m" are the mass of the block.

Explanation

Force due to gravity on the block of mass m is $mg$.

According to the Newton law

$F_{net}=M_{total}\times a$

$F_{net}$is the net force on mass m.

$M_{total}$is the total mass of both the blocks.

a is the acceleration of the mass m.

localid="1649252977169" $$\begin{gathered} mg=\left(M+m\right)a \\ a=\frac{mg}{M+m} \end{gathered}$$

If the block mass of m is moving downwards then the tension in the spring is

localid="1649253032356" $T=mg-ma\cdots \cdots \left(1\right)$

According to the Newton law

$F_{net}=M_{total}\times a$

where $F_{net}$ and a is the net force and acceleration of the mass m respectively.

$$\begin{gathered} mg=\left(M+m\right)a \\ a=\frac{mg}{M+m} \end{gathered}$$

Substitute the value of $a$in the equation $\left(1\right)$

localid="1649253219152" role="math" $$\begin{gathered} T=mg-\frac{m^{2}g}{M+m} \\ T=\frac{mg\left(M+m\right)-m^{2}g}{M+m} \\ T=\frac{Mmg+m^{2}g-m^{2}g}{M+m} \\ T=\frac{Mmg}{M+m} \end{gathered}$$

The tension in the string is $T=\frac{Mmg}{M+m}$