91Ó°ÊÓ

The force $F$can be expressed in terms of acceleration $a$and the mass $m$of the object as follows:

$F=ma$

Let us consider the number of rubber bands used to accelerate the object be $n$and the force exerted by each rubber band on the object be $F$.

Now, the total force exerted by the $n$rubber bands is expressed as follows:

$F_n=nF$

The net force on the object due to two rubber bands is,

$F_2=2F$

Since, the two rubber bands causing the object of mass $m$to accelerate at a rate of $a$, replace the force $F_2$with ma.

$ma=2F$

Rearrange the above equation for $a$.

$a=\frac{2F}{m}$

Express the acceleration of the new object with mass $m^{\text{'}}$as follows:

$a^{\text{'}}=\frac{F_n}{m^{\text{'}}}$

Substitute 3a for $a^{\text{'}},\frac{m}{2}$for $m^{\text{'}}$, and nF for $F_n$.

$\begin{array}{l}3a=\frac{nF}{\frac{m}{2}}\\ 3a=\frac{2nF}{m}\end{array}$

Now, replace $a$with $\frac{2F}{m}$and solve for $n$.

$\begin{array}{l}3\left(\frac{2F}{m}\right)=\frac{2nF}{m}\\ n=3\end{array}$

Therefore, the number of rubber bands required is 3.