A 1.00 -kg block initially at rest at the top of a 4.00 -m incline with a slope of \(45.0^{\circ}\) begins to slide down the incline. The upper half of the incline is frictionless, while the lower half is rough, with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.300\). a) How fast is the block moving midway along the incline, before entering the rough section? b) How fast is the block moving at the bottom of the incline?

We are given the following information: - Initial velocity of the block (v₀) = 0 (as the block is at rest) - Mass of the block (m) = 1.00 kg - Length of the incline (L) = 4.00 m - Angle of inclination (θ) = 45° - Coefficient of kinetic friction on the lower half of the incline (µk) = 0.300 We need to find the velocity of the block halfway (v₁) and at the bottom of the incline (v₂).

We need to find the initial gravitational potential energy (PE₀) of the block. This can be calculated using the formula: PE₀ = m * g * h where g is the acceleration due to gravity (9.81 m/s²) and h is the initial height of the block above the base of the incline. To find h, we can use the trigonometric function sine: sin(θ) = h / L h = L * sin(θ) Substitute the given values: h = 4.00 m * sin(45°) h ≈ 2.83 m Now, calculate the initial gravitational potential energy: PE₀ ≈ 1.00 kg * 9.81 m/s² * 2.83 m PE₀ ≈ 27.72 J (joules)

At the midway point, the block has slid down half the inclined plane, and has a potential energy of half the initial potential energy. Since no external force is acting on the block in the first half, the decrease in potential energy is converted to kinetic energy (KE). The formula relating KE and velocity (v) is: KE = 0.5 * m * v^2 At the midway point, PE₁ = 0.5 * PE₀ KE₁ = PE₀ - PE₁ Solving for v₁: v₁ = sqrt((2 * KE₁) / m) Substitute the values and solve for v₁: v₁ ≈ sqrt((2 * (27.72 J * 0.5)) / 1.00 kg) v₁ ≈ 4.2 m/s

Now, we need to find the work done by friction on the block as it slides down the rough section of the incline. The friction force (Ff) acting on the block can be calculated using the formula: Ff = µk * Fn where Fn is the normal force acting on the block. Due to the inclined plane, the normal force will be: Fn = m * g * cos(θ) Substitute the given values to find Ff: Ff ≈ 0.300 * (1.00 kg * 9.81 m/s² * cos(45°)) Ff ≈ 2.07 N The work done by friction (Wf) on the block can be calculated using the formula: Wf = Ff * d where d is the distance traveled by the block in the rough section. Since the rough section is half the length of the inclined plane, d = L / 2. Substitute the values: Wf ≈ 2.07 N * (4.00 m / 2) Wf ≈ 4.14 J

Finally, we can find the velocity of the block at the bottom of the incline using the conservation of mechanical energy principle. The total mechanical energy (E) at the midway point and the bottom of the incline can be expressed as: E₁ = PE₁ + KE₁ E₂ = PE₂ + KE₂ - Wf Since the block is at the bottom of the incline, its potential energy PE₂ is 0. Therefore, E₂ = KE₂ - Wf Using the conservation of mechanical energy principle (E₁ = E₂), and solving for v₂: v₂ = sqrt((2 * (E₁ + Wf)) / m) Substitute the values and solve for v₂: v₂ ≈ sqrt((2 * (13.86 J + 4.14 J)) / 1.00 kg) v₂ ≈ 5.3 m/s In conclusion, the speed of the block midway along the incline is approximately 4.2 m/s, and the speed of the block at the bottom of the incline is approximately 5.3 m/s.