Question
The student claims that about Jock 3.0 m would have fewer than 96 J of kinetic energy upon a hitting the ground is she correct?
Answers
GPT 3.5
To determine if the student is correct, we need to utilize the equation for kinetic energy:
Kinetic energy (KE) = (1/2) * mass * velocity^2
Since we are given the height (3.0 m) and not the velocity, we need to find the velocity first using the equation for gravitational potential energy:
Potential energy (PE) = mass * gravity * height
Since gravitational potential energy is converted to kinetic energy as the object falls, we can equate the two equations:
PE = KE
mass * gravity * height = (1/2) * mass * velocity^2
height = (1/2) * velocity^2 / gravity
After rearranging the equation, we find:
velocity = √(2 * gravity * height)
Now, let's calculate the velocity:
velocity = √(2 * 9.8 m/s^2 * 3.0 m)
= √(58.8 m^2/s^2)
= 7.67 m/s
Now, let's calculate the kinetic energy at this velocity:
KE = (1/2) * mass * velocity^2
Since we don't have the mass of Jock, we cannot determine the exact kinetic energy. Thus, we cannot say for certain if the student's claim is correct without additional information.
Kinetic energy (KE) = (1/2) * mass * velocity^2
Since we are given the height (3.0 m) and not the velocity, we need to find the velocity first using the equation for gravitational potential energy:
Potential energy (PE) = mass * gravity * height
Since gravitational potential energy is converted to kinetic energy as the object falls, we can equate the two equations:
PE = KE
mass * gravity * height = (1/2) * mass * velocity^2
height = (1/2) * velocity^2 / gravity
After rearranging the equation, we find:
velocity = √(2 * gravity * height)
Now, let's calculate the velocity:
velocity = √(2 * 9.8 m/s^2 * 3.0 m)
= √(58.8 m^2/s^2)
= 7.67 m/s
Now, let's calculate the kinetic energy at this velocity:
KE = (1/2) * mass * velocity^2
Since we don't have the mass of Jock, we cannot determine the exact kinetic energy. Thus, we cannot say for certain if the student's claim is correct without additional information.