A 75.0 kg skydiver is free falling at 65.0 m/s when she pops her parachute. As the 'chute opens she
slows to 5.50 m/s over a distance of 45.0 m. What force does the parachute exert of her harness
during this time?
vf^2=vi^2+2ad where a= force/mass
solve for force
To find the force exerted by the parachute on the skydiver's harness, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.
First, let's calculate the acceleration that the skydiver experiences as she slows down. We can use the equation:
acceleration = (final velocity - initial velocity) / time
Here, the final velocity is 5.50 m/s, the initial velocity is 65.0 m/s, and the time is the distance divided by the average velocity. The average velocity can be calculated by taking the sum of the final and initial velocities and dividing it by 2.
average velocity = (final velocity + initial velocity) / 2
average velocity = (5.50 m/s + 65.0 m/s) / 2 = 35.25 m/s
Now, let's calculate the time:
time = distance / average velocity
time = 45.0 m / 35.25 m/s = 1.2766 s
So, the acceleration is:
acceleration = (5.50 m/s - 65.0 m/s) / 1.2766 s = -51.1914 m/s^2
Now that we have the acceleration, we can find the force:
force = mass × acceleration
The mass is given as 75.0 kg.
force = 75.0 kg × -51.1914 m/s^2
Calculating this value, we find that the force exerted by the parachute on the skydiver's harness during this time is approximately -3839.36 N. The negative sign indicates that the force is directed upward, opposite to the direction of the skydiver's motion.