A 70.0 kg skydiver falls towards the earth. If the force due to air resistance is 500 N, what is the acceleration of the skydiver?
The skydiver opens her chute. The force due to air resistance is now 1200N. What is the acceleration of the skydiver?
M*g = 70*9.8 = 686 N. = Wt. of skydiver.
686-500 = M*a = 70*a.
a =
a = ~2.66 or 2.65714286 / follow henry2's answer, his work checks out.
Well, well, well, isn't the skydiver really going with the flow! Let's solve this together.
In the first scenario, the skydiver is falling towards the earth with a force due to air resistance of 500 N. Now, we know that the force equals mass times acceleration (F = ma). So, 500 N = 70 kg * a. To find the acceleration, we divide both sides by 70 kg, which gives us a = 500 N / 70 kg = 7.14 m/s².
In the second scenario, when the skydiver opens her chute, the force due to air resistance increases to 1200 N. Same deal, my friend! We plug in the numbers: 1200 N = 70 kg * a. Again, divide both sides by 70 kg to find the acceleration: a = 1200 N / 70 kg = 17.14 m/s².
So, in the first scenario, the skydiver is accelerating at 7.14 m/s², and in the second scenario, the acceleration becomes 17.14 m/s². That's quite a "fall-astrophe" difference, right?
To find the acceleration of the skydiver, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In the first scenario, where the force due to air resistance is 500 N, we can write this as:
Net force = mass × acceleration
500 N = 70.0 kg × acceleration
To solve for the acceleration, we rearrange the equation:
acceleration = 500 N / 70.0 kg
acceleration ≈ 7.14 m/s²
So, the acceleration of the skydiver in the first scenario is approximately 7.14 m/s².
Now, let's consider the second scenario, where the force due to air resistance is 1200 N. Using the same formula:
Net force = mass × acceleration
1200 N = 70.0 kg × acceleration
Rearranging the equation to solve for acceleration:
acceleration = 1200 N / 70.0 kg
acceleration ≈ 17.14 m/s²
Therefore, the acceleration of the skydiver after opening her chute and experiencing a force of 1200 N due to air resistance is approximately 17.14 m/s².