Question

D. O8356.675 JE. 5056.506 J2. A bullet of mass 0.0555 kg is fired from a12 m tall building horizontally with a speedof 600 m/s. Calculate its mechanical energyjust after it is fired. (1 point)A. O 4399.788 JB. O 7576.232 JC. 2900.974 JD. O9996.527 JE. O 11970.179 J4

Answer 1

Given:

The mass of the bullet is m = 0.0555 kg

The height of the building is h = 12 m

The speed of the bullet is v = 600 m/s

To find the mechanical energy just after it fired.

Step-by-step explanation:

Mechanical energy is the sum of potential energy and kinetic energy.

The potential energy of the bullet can be calculated by the formula

`P.E.\text{ = mgh}`

Here, g = 9.8 m/s^2 is the acceleration due to gravity.

On substituting the values, the potential energy will be

`\begin{gathered} P.E.\text{ = 0.0555}*9.8*12 \\ =6.5268\text{ J} \end{gathered}`

The kinetic energy can be calculated as

`\begin{gathered} K.E.\text{ = }(1)/(2)mv^2 \\ =(1)/(2)*0.0555*(600)^2 \\ =\text{ 9990 J} \end{gathered}`

Thus, the mechanical energy will be

`\begin{gathered} E=P.E.+K.E. \\ =6.5268+9990 \\ =9996.5268\text{ J} \\ =9996.527\text{ J} \end{gathered}`

Hence, the correct option is D