Question

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.)

Explanation:

Step1: Apply the work - energy theorem

The work - energy theorem states that the work done by non - conservative forces ($W_{nc}$) is equal to the change in mechanical energy ($\Delta E$). Here, the only non - conservative force is the force of kinetic friction. The initial mechanical energy is the kinetic energy $K_{i}=\frac{1}{2}mv_{i}^{2}$ (since there is no potential energy change, assume the surface is horizontal), and the final mechanical energy $K_{f} = 0$ (because the block stops). The work done by friction $W_{f}=-\Delta E_{int}$ (the negative sign is because friction does negative work and the internal energy change is positive). So, $\Delta E_{int}=\frac{1}{2}mv_{i}^{2}$.

Step2: Substitute the given values

Given $m = 3.1\space kg$ and $v_{i}=2.7\space m/s$.
Substitute into the formula $\Delta E_{int}=\frac{1}{2}\times3.1\times(2.7)^{2}$.
First, calculate $(2.7)^{2}=7.29$.
Then, $\frac{1}{2}\times3.1\times7.29 = 1.55\times7.29$.
$1.55\times7.29=(1 + 0.5+0.05)\times7.29=7.29+3.645 + 0.3645$.

Answer:

$11.3$