Calculating Kinetic Energy and Velocity of a Falling Barbell

A. What is the kinetic energy of the barbell as it hits the floor?

A barbel weighing 215N is raised to a height of 2.0 M above the ground. It is then dropped from that height.

B. What is the velocity of the barbell as it hits the floor?

Answer: 430 J 6.26 m/s

Answer:

A. The kinetic energy is the same as the initial potential energy:

PE = mgh = (215 N)(2.0 M) = 430 J

B. The velocity achieved by falling from a height h is given by:

v = √(2gh)

v = √(2·9.8 m/s^2·2 m) = √(39.2 m^2/s^2)

v ≈ 6.26 m/s

When a barbell weighing 215N is raised to a height of 2.0 M above the ground and then dropped, we can calculate its kinetic energy and velocity as it hits the floor. Kinetic energy is the energy possessed by an object due to its motion.

In this case, the potential energy of the barbell at a height of 2.0 M can be calculated using the formula PE = mgh, where PE is the potential energy, m is the mass (215N), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (2.0 M). By substituting the values, we get:

PE = 215 N * 2.0 M = 430 J

Therefore, the kinetic energy of the barbell as it hits the floor is 430 Joules.

To calculate the velocity of the barbell as it hits the floor, we can use the formula v = √(2gh), where v is the velocity, g is the acceleration due to gravity, and h is the height. By substituting the values, we get:

v = √(2 * 9.8 m/s^2 * 2 M) = √(39.2 m^2/s^2) ≈ 6.26 m/s

Therefore, the velocity of the barbell as it hits the floor is approximately 6.26 m/s.