Problem 56 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

Answer: 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.

Step 1: Identify known and unknown variables

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).

Step 2: Calculate gravitational potential energy

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)

Step 3: Find the velocity at the midway point

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

Step 4: Find the work done by friction

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

Step 5: Find the velocity at the bottom of the incline

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.

Key Concepts

Potential Energy

Potential energy is one of the fundamental aspects of physics that deals with the energy that an object possesses because of its position or height. Imagine you hold a book at a certain height. That book has gravitational potential energy because, when released, gravity will act upon it, causing it to fall. The potential energy stored in objects due to their height can be calculated using the formula: \[ PE = m \cdot g \cdot h \] where:

• \( m \) is the mass of the object
• \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \) on Earth)
• \( h \) is the height of the object above the ground or baseline.

When a 1 kg block is placed at the top of an inclined plane, it exhibits potential energy due to its elevated position. As the block begins to slide down, this gravitational potential energy is transformed into kinetic energy, which is the energy of motion. The change in height on an inclined plane can be determined using trigonometry, specifically:

• \( \sin(\theta) = \frac{h}{L} \),

where \( \theta \) is the angle of the incline and \( L \) is the length of the incline. Observing the block at the top of the incline makes it evident that potential energy is fundamentally linked to height.

Friction

Friction is the resistive force that acts against the motion of an object sliding along a surface. It's the reason why sliding on a floor isn't as smooth as on an ice rink. Friction occurs due to the interactions between the surfaces at the microscopic level, and it can significantly affect the energy of moving objects.
Two types of friction mainly concern us: static and kinetic. In this case, the focus is on kinetic friction because once the block on the inclined plane starts moving, it's kinetic energy that comes into play.
The kinetic frictional force (\( F_f \)) can be calculated using:\[ F_f = \mu_k \cdot F_n \] where:

• \( \mu_k \) is the coefficient of kinetic friction
• \( F_n \) is the normal force, which is the force perpendicular to the surface

On the rough section of the incline, friction does work on the block and reduces its mechanical energy. The work done by friction (\( W_f \)) can be calculated by multiplying frictional force by the distance over which it acts. Friction does negative work because it opposes motion, thus subtracting from the block’s kinetic energy as it moves down the incline, requiring us to account for this loss when determining final speeds.

Inclined Plane

An inclined plane simplifies the study of motion and forces. It’s basically a flat surface tilted at an angle to the horizontal. In real-life applications, it helps in moving heavy objects with less force. In physics problems, understanding inclined planes aids in analyzing motion and forces because it introduces both the concepts of gravity and friction.
When you place an object on an inclined plane, gravity acts on it, pulling it downwards and along the slope. However, the force of gravity is not working directly downward, as it would with a flat surface, but instead works at an angle, necessitating us to consider both components of the force - parallel and perpendicular.

• The component of gravitational force pulling the object down the plane is \( m \cdot g \cdot \sin(\theta) \).
• The normal force, which affects friction, is calculated with \( m \cdot g \cdot \cos(\theta) \).

This breakdown is crucial when analyzing speed changes on different sections of the plane, as it impacts how forces balance and work on the object. By understanding how these interact on an incline, you should better grasp the concepts of energy transformation and motion.