Question

tep 6: calculate the pred
bottles
this part of the experiment, you will be chan
eed of the bottle by dropping it from differer
ou will use the same mass, 0.250 kg, for eac
cord this mass in table b for each velocity.
alculate the expected kinetic energy (ke) at
elocity. use the formula ( ke = \frac{1}{2}mu^2 ),
where m is the mass and v is the speed. record your
alculations in table b of your student guide.
of the falling soda
a bottle is 2 m/s, the ke is
a bottle is 3 m/s, the ke is
when the speed of the bottle is 4 m/s, the ke is
kg • m²/s².
when the speed of the bottle is 5 m/s, the ke is
kg • m²/s².
when the speed of the bottle is 6 m/s, the ke is
kg • m²/s².

Explanation:

Step1: Identify formula and values

The formula for kinetic energy is \( KE = \frac{1}{2}mu^2 \), where \( m = 0.250 \, \text{kg} \) and we'll calculate for \( u = 4 \, \text{m/s} \), \( u = 5 \, \text{m/s} \), \( u = 6 \, \text{m/s} \).

Step2: Calculate KE for \( u = 4 \, \text{m/s} \)

Substitute \( m = 0.250 \) and \( u = 4 \) into the formula:
\( KE = \frac{1}{2} \times 0.250 \times (4)^2 \)
\( = \frac{1}{2} \times 0.250 \times 16 \)
\( = 0.125 \times 16 \)
\( = 2 \, \text{kg·m}^2/\text{s}^2 \)

Step3: Calculate KE for \( u = 5 \, \text{m/s} \)

Substitute \( m = 0.250 \) and \( u = 5 \):
\( KE = \frac{1}{2} \times 0.250 \times (5)^2 \)
\( = \frac{1}{2} \times 0.250 \times 25 \)
\( = 0.125 \times 25 \)
\( = 3.125 \, \text{kg·m}^2/\text{s}^2 \)

Step4: Calculate KE for \( u = 6 \, \text{m/s} \)

Substitute \( m = 0.250 \) and \( u = 6 \):
\( KE = \frac{1}{2} \times 0.250 \times (6)^2 \)
\( = \frac{1}{2} \times 0.250 \times 36 \)
\( = 0.125 \times 36 \)
\( = 4.5 \, \text{kg·m}^2/\text{s}^2 \)

Answer:

• When speed is 4 m/s, KE is \( 2 \, \text{kg·m}^2/\text{s}^2 \)
• When speed is 5 m/s, KE is \( 3.125 \, \text{kg·m}^2/\text{s}^2 \)
• When speed is 6 m/s, KE is \( 4.5 \, \text{kg·m}^2/\text{s}^2 \)