91影视

The object鈥檚 acceleration can be identified by evaluating the value of the applied force and the mass of the object. Its value is directly related to the value of the applied force.

Applying Newton鈥檚 second law for a suspended ball accelerating in the upward direction,

$F_T-mg=ma$.

Here, $F_T$is the tension force, $m$is the mass, $a$is the acceleration, and $g$is the acceleration due to gravity.

The above equation can be written as

$\begin{array}{c}{F}_T=ma+mg\\ F_T=m\left(a+g\right)\end{array}$

The link will break only when the applied force is greater than 22.2 N. So, the range of the force where the elevator can accelerate is given as follows:

$m\left(a+g\right)>22.2\mathrm{N}$will cause the fishing line to break.

$m\left(a+g\right)<22.2\mathrm{N}$will leave the fishing line unbroken.

The highest mass among the five balls to keep the line unbroken is 2.05 kg, and the lowest mass is 2.00 kg.

Thus, the range of acceleration is

$\begin{array}{c}\left(a+g\right)<\frac{22.2\mathrm{N}}{2\mathrm{kg}}\\ a<\left(11.1\mathrm{m}{/\mathrm{s}}^2\right)-g\end{array}$

and

$\begin{array}{c}\left(a+g\right)>\frac{22.2\mathrm{N}}{2.05\mathrm{kg}}\\ a>\left(10.829\mathrm{m}{/\mathrm{s}}^2\right)-g\end{array}$

The range can be calculated as

$\begin{array}{c}\left(10.829\mathrm{m}{/\mathrm{s}}^2\right)-g<a<\left(11.1\mathrm{m}{/\mathrm{s}}^2\right)-g\\ \left(10.829\mathrm{m}{/\mathrm{s}}^2\right)-\left(9.8\mathrm{m}{/\mathrm{s}}^2\right)<a<\left(11.1\mathrm{m}{/\mathrm{s}}^2\right)-\left(9.8\mathrm{m}{/\mathrm{s}}^2\right)\\ \left(1.029\mathrm{m}{/\mathrm{s}}^2\right)<a<\left(1.3\mathrm{m}{/\mathrm{s}}^2\right)\end{array}$

Thus, the elevator can accelerate between $1.029\mathrm{m}{/\mathrm{s}}^2$and $1.3\mathrm{m}{/\mathrm{s}}^2$.