You lower a bundle weighing 64.0 kg to the ground with a rope that will snap if the tension exceeds 578.2 N. If the height is 8.0 m and you barely avoid snapping the rope, what speed will the bundle hit the ground?
Tension=mg-ma
solve for a. Then, speed equals
Vf^2=2a*8
I still don't get it. I got 3.5 which is not the right answer
nevermind the answer is 3.50
To find the speed at which the bundle hits the ground, we can use the principle of conservation of energy.
First, we need to determine the potential energy of the bundle at the height of 8.0 m. The potential energy is given by the formula:
Potential Energy = mass * gravity * height
where
mass = 64.0 kg (bundle's weight)
gravity = 9.8 m/s² (acceleration due to gravity)
height = 8.0 m
Substituting the values into the formula, we have:
Potential Energy = 64.0 kg * 9.8 m/s² * 8.0 m
Next, we'll equate the potential energy to the maximum tension the rope can handle. Since the rope is about to snap, the potential energy is equal to the maximum tension.
Potential Energy = Maximum Tension = 578.2 N
Now, we can calculate the kinetic energy of the bundle just before it hits the ground using the formula:
Kinetic Energy = (1/2) * mass * velocity²
where
mass = 64.0 kg (bundle's weight)
velocity = unknown speed
Substituting the values, we have:
Kinetic Energy = (1/2) * 64.0 kg * velocity²
Since energy is conserved, the potential energy is equal to the kinetic energy:
Potential Energy = Kinetic Energy
Therefore, we can set up the equation:
(1/2) * 64.0 kg * velocity² = 578.2 N
Simplifying the equation:
32.0 kg * velocity² = 578.2 N
Now, let's solve for the velocity. Dividing both sides of the equation by 32.0 kg, we get:
velocity² = 578.2 N / 32.0 kg
velocity² = 18.069 m²/s²
Taking the square root of both sides, we find:
velocity = √18.069 m/s
Thus, the speed at which the bundle hits the ground is approximately 4.25 m/s.