Question

The hammer throw is a track and field event in which a 8.34- kg ball (the "hammer"), starting from rest, is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion. In one throw, the hammer is given a speed of 25.0 m/s. For comparison, a . 22 caliber bullet has a mass of about 2.88 g and, Starting from rest, exits the barrel of a gun with a speed of 440 m/s. Determine the work done to launch the motion of (a) the hammer and (b) the bullet. (a) Number Units (b) Number Units

Answer 1

The work done to launch the hammer and bullet is found using the kinetic energy formula, resulting in 2606.25 joules for the hammer and 278.784 joules for the bullet.

Step-by-step explanation:

To calculate the work done to launch the hammer and the bullet, we can use the formula for work related to kinetic energy:

Work Done (W) = Kinetic Energy (KE) = ½ * mass (m) * velocity (v)^2

For the hammer:

W = ½ * 8.34 kg * (25.0 m/s)^2

W = ½ * 8.34 kg * 625 m^2/s^2

W = 2606.25 J (joules)

For the bullet:

Since the mass is given in grams, we first convert it to kilograms: 2.88 g = 0.00288 kg.

W = ½ * 0.00288 kg * (440 m/s)^2

W = ½ * 0.00288 kg * 193600 m^2/s^2

W = 278.784 J (joules)

In conclusion, the work done to launch the hammer is 2606.25 J and the work done to launch the bullet is 278.784 J.