Question

The hammer throw is a track-and-field event in which a 7.3 kg ball (the “hammer”), starting from rest, is whirled around in a circle sev- eral 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 29 m/s. For comparison, a .22 caliber bullet has a mass of 2.6 g and, start- ing from rest, exits the barrel of a gun at a speed of 410 m/s. Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

Answer 1

Answer:

Part a)

W = 3070 J

Part b)

W = 218.5 J

Explanation:

Part a)

Work done to throw the ball upwards in the sky is given by work energy theorem

It is given as

$W = \frac{1}{2}mv^2 - 0$

$W = \frac{1}{2}(7.3)(29^2)$

$W = 3070 J$

Part b)

Now  similarly we have to find the work done to launch the bullet from gun is also given by change in kinetic energy

$W = \frac{1}{2}mv^2 - 0$

$W = \frac{1}{2}(0.0026)(410^2)$

$W = 218.5 J$