Question

which most simplified form of the law of conservation of energy describes the motion of the block as it slides on the floor from the moment it stops? view available hint(s)
$\boldsymbol{\frac{1}{2}mv_i^2 + mgh_i + w_{\text{nc}} = \frac{1}{2}mv_f^2 + mgh_f}$
$\boldsymbol{\frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2}$
$\boldsymbol{\frac{1}{2}mv_i^2 + w_{\text{nc}} = 0}$
$\boldsymbol{\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f}$
$\boldsymbol{\frac{1}{2}mv_i^2 + mgh_i + \frac{1}{2}kx_i^2 + w_{\text{nc}} = \frac{1}{2}mv_f^2 + mgh_f + \frac{1}{2}kx_f^2}$
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Explanation:

1. Analyze the scenario: The block slides on the floor until it stops. We consider the initial and final states.

• Initial state: The block has initial kinetic energy $\frac{1}{2}mv_{i}^{2}$ (assuming it has some initial velocity) and initial gravitational potential energy $mgh_{i}$. But when sliding on the floor (horizontal surface), the height doesn't change, so $h_{i} = h_{f}$, which means the gravitational potential energy terms will cancel out. Also, there is no spring involved here (so no $\frac{1}{2}kx^{2}$ terms).
• Final state: The block comes to rest, so final kinetic energy $\frac{1}{2}mv_{f}^{2}=0$. The work done by non - conservative forces (like friction) is $W_{nc}$.
• The general work - energy theorem (conservation of energy with non - conservative work) is $\frac{1}{2}mv_{i}^{2}+mgh_{i}+W_{nc}=\frac{1}{2}mv_{f}^{2}+mgh_{f}$. But since $h_{i} = h_{f}$ (horizontal motion) and $v_{f} = 0$, we can simplify this.
• Substituting $v_{f}=0$ and $h_{i}=h_{f}$ into the general formula: $\frac{1}{2}mv_{i}^{2}+mgh_{i}+W_{nc}=\frac{1}{2}(0)v_{f}^{2}+mgh_{i}$. The $mgh_{i}$ terms cancel out, and we get $\frac{1}{2}mv_{i}^{2}+W_{nc}=0$.
• Let's check other options:
• Option $\frac{1}{2}mv_{i}^{2}+mgh_{i}+W_{nc}=\frac{1}{2}mv_{f}^{2}+mgh_{f}$: This is the general formula, but we can simplify it further as the height doesn't change and final velocity is zero.
• Option $\frac{1}{2}mv_{i}^{2}=\frac{1}{2}mv_{f}^{2}$: This would be true only if there is no non - conservative work (like friction), but when sliding to a stop, friction (non - conservative force) is doing work.
• Option $\frac{1}{2}mv_{i}^{2}+mgh_{i}=\frac{1}{2}mv_{f}^{2}+mgh_{f}$: This ignores the non - conservative work done by friction, which is present when the block slides to a stop.
• Option $\frac{1}{2}mv_{i}^{2}+mgh_{i}+\frac{1}{2}kx_{i}^{2}+W_{nc}=\frac{1}{2}mv_{f}^{2}+mgh_{f}+\frac{1}{2}kx_{f}^{2}$: There is no spring involved in the problem (sliding on the floor), so the spring potential energy terms ($\frac{1}{2}kx^{2}$) are not applicable.

Answer:

$\boldsymbol{\frac{1}{2}mv_{i}^{2}+W_{nc}=0}$ (the third option in the list)