Question

Compute the gravitational potential energy of a water bottle – you choose its size – that is lifted 1689 centimeters into the air near the surface of the Earth. Then compute how much kinetic energy is released if the water bottle is dropped and it falls one meter. Give your answers in units of joules.

Answer 1

a. The gravitational potential energy of the water bottle is 164.522 J

b. The final kinetic energy of bottle after dropping one meter is 9.8 J

a. Gravitational potential energy

The gravitational potential energy of the water bottle is 164.522 J

The gravitational potential energy of the water bottle is U = mgh where

• m = mass of water bottle = 1 kg,
• g = acceleration due to gravity = 9.8 m/s² and
• h = height of water bottle = 1689 cm = 16.89 m

So, U = mgh

= 1 kg × 9.8 m/s² × 16.89 m

= 165.522 kgm²/s²

= 164.522 J

The gravitational potential energy of the water bottle is 164.522 J

b. Conservation of energy

The final kinetic energy of bottle after dropping one meter is 9.8 J

From the law of conservation of energy, the gravitational potential energy change when the bottle drops one meter, ΔU equals the kinetic energy change in kinetic energy, ΔK

ΔU = -ΔK

mgΔh = -(K' - K) where

• m = mass of water bottle = 1 kg,
• g = acceleration due to gravity = 9.8 m/s²,
• Δh = height drop = - 1 m (negative since the height drops),
• K = initial kinetic energy of water bottle = 0 J(since it starts from rest) and
• K' = final kinetic energy of bottle after dropping one meter.

Final kinetic energy of bottle after dropping one meter.

Making K' subject of the formula, we have

mgΔh = -(K' - K)

mgΔh = -K' + K

K' = K - mgΔh

Substituting the values of the variables into the equation, we have

K' = K - mgΔh

K' = 0 J -  1 kg × 9.8 m/s² × (- 1 m)

K' = 0 J + 9.8 kgm²/s²

K' = 0 J + 9.8 J

K' = 9.8 J

The final kinetic energy of bottle after dropping one meter is 9.8 J

Learn more about gravitational potential energy here:

https://brainly.com/question/12314347