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

Answer:

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

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 :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 12\ kg\times (20\ m/s)^2$

K = 2400 J

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

P = mgh

$P=12\ kg\times 9.8\ m/s^2\times 18\ m$

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.