Question

The strings of a parachute can bear a maximum tension of $75{\text{ }}kg$ -weight. By what minimum acceleration can a person of $96{\text{ }}kg$ descend by means of this parachute?

Answer 1

To solve the topic, we will use the formula to determine the minimal acceleration at which a body of a particular mass will descend using a parachute. When a person's weight is a factor in how they move. The calculation should be downward force minus upward force because the net force in the question is downward. As a result, there is a downward net force.

Formula used:
$mg-T = ma$
Here, the tension in the string is $\left( T \right)$.
Mass of the man is $\left( m \right)$.
Acceleration due gravity $\left( g \right)$.

Complete step by step answer:
In the above question it is given to us that;
The tension in the string is $\left( T \right) = 75{\text{ }}kg$.
Mass of the man is $\left( m \right) = 96\,kg$.
Acceleration due gravity $\left( g \right) = 9.8\,m{s^{ - 1}}$.
Now, as we know that;
$mg - T = ma$
Therefore from this equation that we use when weight of a body plays a role in it’s motion we will find the value of $'a'$ by putting all the given values and equating them.
$a = \dfrac{{mg - T}}{m}$
Now, putting all the given value we will evaluate it as;
$a = \dfrac{{96 \times 9.8 - 75 \times 9.8}}{{96}}$
On further evaluation
$\Rightarrow a = \dfrac{{940.8 - 735}}{{96}}$

$\Rightarrow a = \dfrac{205.8}{96}$
After the final evaluation we will have our answer as
$\therefore a = 2.14\,m{s^{ - 2}}$
Hence, the Person should descend by acceleration is $2.14\,m{s^{ - 2}}$.

Note: If a body is suspended on a string and moves upwards at a constant rate, then net force Equals$ma$ . When weight is not a factor in motion, $T = {\text{ }}ma$ equals upward force $T$ minus downward force $mg$ . If a body is on a horizontal floor and is being dragged by a string with tension $T$ , and the body's acceleration is $a$ , weight acting vertically downwards has no effect on horizontal motion. As a result, $T = ma$ will be the equation.