Answer:
0.137 J
Explanation:
The conservation of energy states that the mechanical energy of the system is conserved:
[tex]E=U+K=const.[/tex]
where
[tex]U=mgh[/tex] is the potential energy, where m is the mass, g is the gravitational acceleration and h the height above the ground
[tex]K=\frac{1}{2}mv^2[/tex] is the kinetic energy, where m is the mass and v is the speed
- When the ball is dropped, the kinetic energy is zero (because it starts from rest). So, the mechanical energy is just potential energy:
[tex]E=U=mgh=(0.028 kg)(9.8 m/s^2)(1.0 m)=0.274 J[/tex]
- When it is h=0.5 m above the ground, part of the energy is now kinetic energy. The potential energy now is
[tex]U=mgh=(0.028 kg)(9.8 m/s^2)(0.5 m)=0.137 J[/tex]
And since the mechanical energy is conserved, we can write
[tex]E=U+K\\K=E-U=0.274 J-0.137 J=0.137 J[/tex]
So, the kinetic energy is 0.137 J.
