Question

part 8 of 9 - analyze
the friends consider a block of mass 3.1 kg set in motion by an external force. the initial velocity is 2.7 m/s, and the coefficient of kinetic friction is 0.01. what do they find as the final change in internal energy of the system once the block comes to a complete stop? (enter your answer in j.)
11.3 11.3 j
going further: effect of doubling the
mass
jessie and shawna now consider a block moving with an initial velocity, ( v_i ), on a rough surface, and coming to a complete stop. although the simulation does not allow them to change the mass of the block, shawna imagines running the experiment again, this time with the mass doubled. how should she say doubling the mass affects the change in internal energy of the block - surface system?

• the change in internal energy doubles.
• the change in internal energy is the same.
• the change in internal energy is half.
• the change in internal energy quadruples.

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Explanation:

Step1: Determine the initial kinetic energy formula

The initial kinetic energy \(K_{i}=\frac{1}{2}mv^{2}\). When the block stops, all of its initial kinetic energy is converted into internal energy (due to friction). So, \(\Delta E_{int}=K_{i}=\frac{1}{2}mv^{2}\)

Step2: Analyze the effect of doubling the mass

Let the initial mass be \(m\) and the new mass be \(m' = 2m\). The initial velocity \(v\) remains the same.

For the original mass \(m\), \(\Delta E_{int1}=\frac{1}{2}mv^{2}\)

For the new mass \(m'=2m\), \(\Delta E_{int2}=\frac{1}{2}(2m)v^{2}\)

Simplify \(\Delta E_{int2}\): \(\Delta E_{int2}= 2\times(\frac{1}{2}mv^{2})\)

Since \(\Delta E_{int1}=\frac{1}{2}mv^{2}\), we have \(\Delta E_{int2} = 2\Delta E_{int1}\)

Answer:

A. "The change in internal energy doubles."