Question

data table kinetic energy
rudolf and dasher wanted to test how velocity affects kinetic energy.
rudolf thinks that if a toy sleigh has more velocity it will increase the
amount of kinetic energy. but dasher thinks that you cant really predict
kinetic energy, it just happens randomly.
to figure out who was right they made an experiment. they set up a toy
track and used a stopwatch to see how fast the car was going. they let the
toy sleigh go down the track and recorded the speed each time. they
repeated this 6 times changing the speed for each trial. once they got all
their results they used the following equation to find kinetic energy.
ke = 1/2 mv^2 they entered the amount of kinetic energy in the table below.
multiple choice 1 point
which of the following hypotheses would best match what rudolf thinks?
if velocity is decreased then kinetic energy will increase
if kinetic energy is increased then it is because of a random force
if mass is increased then kinetic energy will decrease
if velocity is increased then kinetic energy will increase

Explanation:

Step1: Analyze the relationship between velocity and kinetic energy

The formula for kinetic energy is \(KE = \frac{1}{2}mv^{2}\). From the data table, when the mass \(m = 2kg\) (constant in all trials), as the velocity \(v\) increases, the kinetic energy \(KE\) increases. For example, when \(v = 1m/s\), \(KE=\frac{1}{2}\times2\times1^{2}=1\) joule; when \(v = 2m/s\), \(KE=\frac{1}{2}\times2\times2^{2} = 4\) joules.

Step2: Evaluate each hypothesis

• If velocity is decreased then kinetic energy will increase: This is incorrect. As per the formula \(KE=\frac{1}{2}mv^{2}\), when \(m\) is constant, \(KE\) is directly proportional to \(v^{2}\). Decreasing \(v\) will decrease \(KE\).
• If kinetic energy is increased then it is because of a random force: This is incorrect. The data shows a clear relationship between \(v\) (and \(m\)) and \(KE\) based on the formula \(KE=\frac{1}{2}mv^{2}\), not random forces.
• If mass is increased then kinetic energy will decrease: This is incorrect. From \(KE=\frac{1}{2}mv^{2}\), when \(v\) is constant, increasing \(m\) will increase \(KE\) (e.g., if \(v = 1m/s\) and \(m = 3kg\), \(KE=\frac{1}{2}\times3\times1^{2}=1.5\) joules, which is more than when \(m = 2kg\) and \(v = 1m/s\) (\(KE = 1\) joule)).
• If velocity is increased then kinetic energy will increase: This is correct. Since \(KE=\frac{1}{2}mv^{2}\) and when \(m\) is constant (as in the trials where \(m = 2kg\)), increasing \(v\) (e.g., from \(v = 1m/s\) to \(v = 6m/s\)) leads to an increase in \(KE\) (from \(1\) joule to \(36\) joules).

Answer:

If velocity is increased then kinetic energy will increase.