Question

A 12-kg projectile is launched with an initial vertical speed of 20 m/s. It rises to a maximum height of 18 m above the launch point. What is the change in mechanical energy caused by the dissipative (air) resistive force on the projectile during this ascent

Answer 1

Change in mechanical energy,
`\Delta E=283.2\ J`

Step-by-step explanation:

It is given that,

Mass of the projectile, m = 12 kg

Speed of the projectile, v = 20 m/s

Maximum height, h = 18 m

Initially, the projectile have only kinetic energy. it is given by :

`K=(1)/(2)mv^2`

`K=(1)/(2)* 12\ kg* (20\ m/s)^2`

K = 2400 J

Finally, it have only potential energy. it is given by :

P = mgh

`P=12\ kg* 9.8\ m/s^2* 18\ m`

P =2116.8 J

The change in mechanical energy is given by :

`\Delta E=K-P`

`\Delta E=2400-2116.8`

`\Delta E=283.2\ J`

So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.