A 6.0 -kg block, starting from rest, slides down a frictionless incline of length \(2.0 \mathrm{m} .\) When it arrives at the bottom of the incline, its speed is \(v_{\mathrm{f}} .\) At what distance from the top of the incline is the speed of the block \(0.50 v_{\mathrm{f}} ?\)

Since the block is sliding down a frictionless incline, let's start by using the conservation of mechanical energy principle. The mechanical energy of the block at the top and bottom of the incline will be equal. At the top of the incline, the block has only potential energy (PE) due to its height \(h\), which can be calculated as \(PE = mgh\), where \(m\) is the mass of the block, \(g\) is the acceleration due to gravity, and \(h\) is the height of the incline. At the bottom of the incline, the block has only kinetic energy (KE) due to its motion, calculated as \(KE = \frac{1}{2}mv_f^2\), where \(v_f\) is the final speed of the block at the bottom of the incline. The energy conservation equation will be: \(\begin{aligned} mgh = \frac{1}{2}mv_f^2 \end{aligned}\)

We don't know the height of the incline, but we can find the relationship between the height and the length of the incline. Let \(\theta\) be the angle between the incline and horizontal. Then, \(\begin{aligned} h = (2.0 \mathrm{m})\sin(\theta) \end{aligned}\) Now, using the energy conservation equation, we get: \(\begin{aligned} mgh = \frac{1}{2}mv_f^2 \end{aligned}\) \(\begin{aligned} g(2.0 \mathrm{m})\sin(\theta) = v_f^2 \end{aligned}\) We can solve for \(v_f\).

Now, we want to find the speed of the block when it is half its final speed \(v = 0.50 v_f\). To find this, we will use the kinematic equation: \(\begin{aligned} v^2 = v_i^2 + 2as \end{aligned}\) where \(v_i\) is the initial speed (0 in this case), \(a\) is the acceleration down the incline, \(s\) is the distance traveled, and \(v\) is the final speed. Since \(v = 0.50 v_f\), we can rewrite the equation as: \(\begin{aligned} (0.50 v_f)^2 = 0 + 2as \end{aligned}\)

Now, we want to find the distance \(s\) when the block reaches half its final speed. From Step 2, we have: \(\begin{aligned} g(2.0 \mathrm{m}) \sin(\theta) = v_f^2 \end{aligned}\) From our kinematic equation in Step 3: \(\begin{aligned} (0.50 v_f)^2 = 0 + 2as \end{aligned}\) Plug the expression for \(v_f^2\) into the equation: \(\begin{aligned} (0.50)^2 g(2.0 \mathrm{m}) \sin(\theta) = 2as \end{aligned}\) Now, we can solve for \(s\): \(\begin{aligned} s = \frac{1}{4}(2.0 \mathrm{m}) \sin(\theta) = 0.50 \mathrm{m} \sin(\theta) \end{aligned}\) So, at a speed of half of its final speed, the block will be \(0.50 \mathrm{m} \sin(\theta)\) away from the top of the incline.