Question

A ball of mass 0.2 kg is dropped from a height of 10 m. How much mechanical energy does it have right before it hits the ground? (Assume there is no air resistance.) Acceleration due to gravity is g 9.8 m/s2.

(PLEASE HELP NEED ANSWERS ASAP)

Answer 1

0.2 kg * 9.81 m/s^2* 10= 19.62 or B

Answer 2

19.6 J

Step-by-step explanation:

Hello

The law of conservation of energy states that the total amount of energy in any isolated physical system remains unchanged over time, although this energy can be transformed into another form of energy, in this case we initially have gravitational potential energy (due to height and gravity) and then it is transformed into kinetic energy (due to the movement)

`E_(g) =mgh`

where m is the mass , g the gravity and h the height

`E_(k) =(m*v^(2) )/(2)`

assuming that there is no air resistance the energy is conserved

`E_(g) =E_(k) \\mgh = (m*v^(2) )/(2) \\\\`

`E_(g) =mgh = 0.2 kg*9.8(m)/(s^(2) ) *10 m\\\\E_(g)=19.6 Joules\\ E_(g)=E_(k)=19.6 J\\`

the energy it has right before it hits the ground is 19.6 J

I hope it helps.