A 75 kg person escapes from the burning building by jumping from the window 25 m above a catching net air resistance exerts a 95 N force on the person during the fall apply Newton second law find the acceleration during the fall find the velocity and free body diagram
To find the acceleration during the fall, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's identify the forces acting on the person during the fall:
1. Weight (mg): The weight of the person can be calculated by multiplying their mass (m) by the acceleration due to gravity (g).
Weight = m * g
2. Air resistance (F_air): The problem states that air resistance exerts a force of 95 N on the person during the fall.
According to Newton's second law, the net force on the person is the sum of the weight and the air resistance:
Net force = Weight - Air resistance
Since the person is falling vertically downward, the net force is equal to the mass of the person multiplied by the acceleration:
Net force = m * a
Now, we can equate the two expressions for the net force:
m * a = Weight - Air resistance
Substituting the known values:
75 kg * a = (75 kg * g) - 95 N
To find the acceleration (a), we need to know the value of the acceleration due to gravity (g). On Earth, this value is approximately 9.8 m/s^2.
Plugging in the value of g:
75 kg * a = (75 kg * 9.8 m/s^2) - 95 N
Simplifying the equation:
75 kg * a = 735 kg*m/s^2 - 95 N
75 kg * a = 735 kg*m/s^2 - 95 kg*m/s^2 (since 1 N = 1 kg*m/s^2)
Combining the terms with the same units:
75 kg * a = 640 kg*m/s^2
Finally, we can solve for the acceleration:
a = (640 kg*m/s^2) / 75 kg = 8.53 m/s^2
Therefore, the acceleration during the fall is approximately 8.53 m/s^2.
To find the velocity, we can use the equation of motion:
v^2 = u^2 + 2aS
Where:
- v is the final velocity
- u is the initial velocity (which is 0 m/s as the person is released)
- a is the acceleration during the fall (which we found to be 8.53 m/s^2)
- S is the displacement (25 m in this case, as the person falls 25 m)
Plugging in the known values:
v^2 = 0^2 + 2 * 8.53 m/s^2 * 25 m
Simplifying the equation:
v^2 = 2 * 213.25 m^2/s^2
v^2 = 426.5 m^2/s^2
Taking the square root of both sides to find the velocity:
v = √(426.5 m^2/s^2) ≈ 20.65 m/s
Therefore, the velocity of the person just before reaching the catching net is approximately 20.65 m/s.
Here is the free body diagram for the person during the fall:
+---------+
| |
| |
Weight <---> |
(mg) | |
| O |
Air | |
resistance (F_air) |
| |
| |
+---------+
In this diagram, the weight (mg) is acting downwards, while the air resistance (F_air) is acting upwards. The net force is the difference between these two forces and causes the person to accelerate downwards.