Problem 16 - University Physics with Modern Physics

Question

A 8.00 -kg block of ice, released from rest at the top of a \(1.50-\mathrm{m}\)-long frictionless ramp, slides downhill, reaching a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 \(\mathrm{N}\) parallel to the surface of the ramp?

Step-by-Step Solution

Answer

(a) The angle is approximately 12.3°. (b) The speed at the bottom is 1.93 m/s.

Step 1: Define Known Variables

We start by listing the given information. The mass of the block of ice, \( m \), is 8.00 kg. The length of the ramp, \( L \), is 1.50 m. The final speed at the bottom, \( v_f \), is 2.50 m/s. We need to find the angle \( \theta \) for part (a).

Step 2: Apply Conservation of Energy (Part a)

For the frictionless scenario, we apply the conservation of mechanical energy. Initially, the ice block has gravitational potential energy at the top, which converts completely to kinetic energy at the bottom.\[ mgh = \frac{1}{2}mv_f^2\]Where \( h \) is the vertical height, which can be related to the length of the ramp \( L \) using \( h = L \sin\theta \). Solve for \( h \) first.

Step 3: Calculate Vertical Height (Part a)

From the conservation of energy equation, substitute the known values:\[ 8.00 \times 9.8 \times h = \frac{1}{2} \times 8.00 \times (2.50)^2\]Solving this gives \( h = 0.3189 \) m.

Step 4: Find the Angle Theta (Part a)

Using the relation \( h = L \sin\theta \), substitute the known values:\[ 0.3189 = 1.50 \sin\theta\]Solving for \( \sin\theta \) gives \( \sin\theta = 0.2126 \), so \( \theta = \arcsin(0.2126) \approx 12.3^\circ \).

Step 5: Handle Friction (Part b)

For part (b), we include the work done against friction. The friction force \( f \) is 10.0 N, opposing the motion. The work done by the friction force is: \[ W_{friction} = -f \cdot L = -10.0 \times 1.50 = -15.0 \text{ J} \]

Step 6: Modify Energy Equation for Friction (Part b)

Apply the work-energy principle considering the work done against friction:\[ mgh - 15.0 = \frac{1}{2} m v_f^2\]Substitute \( h = 0.3189 \) m obtained earlier.

Step 7: Solve for Final Speed with Friction (Part b)

Set up the energy equation:\[ 8.00 \times 9.8 \times 0.3189 - 15.0 = \frac{1}{2} \times 8.00 \times v_f^2\]This simplifies to solve for \( v_f \), yielding \( v_f \approx 1.93 \text{ m/s} \).

Key Concepts

Conservation of Energy

The conservation of energy principle is a cornerstone of physics. It states that energy cannot be created or destroyed; it can only be transformed from one form to another. In this exercise, we consider the ice block at rest at the top of the ramp as it has gravitational potential energy. Once it slides down the ramp, this potential energy is converted into kinetic energy.

For a frictionless scenario, we can use the formula:

• Potential energy at the top: \( mgh \)
• Kinetic energy at the bottom: \( \frac{1}{2}mv_f^2 \)

These energies equal each other according to the conservation of energy principle, giving us \( mgh = \frac{1}{2}mv_f^2 \). Notice how the mass cancels out, which highlights that the change in energy and consequently speed is independent of the mass of the block. This transforms a potentially mind-boggling concept into something quite simple and beautiful in its universality.

Understanding this energy transformation helps us calculate the height \( h \) of the ramp and ultimately find the angle of the incline.

Newton's Laws of Motion

Newton's laws provide the framework for understanding motion and forces, especially when multiple forces like gravity and friction come into play. In this exercise, Newton's second law of motion is central. It tells us that an object’s change in motion (acceleration) is directly proportional to the net force acting upon it and inversely proportional to its mass \( (F = ma) \). However, here, since we're working with energy rather than directly looking at forces and accelerations, we see an indirect application of the law.

When the block slides down the ramp, gravity is the force driving its motion. If friction were present, it would oppose this force, slowing down the block. Recognizing this distinction provides insight into how forces balance and interact, which is invaluable for tackling complex physics problems.

Friction in Physics

Friction is the force resisting the relative motion of solid surfaces sliding against each other. Even though this specific problem has a frictionless scenario initially, part (b) explores how friction affects the block’s speed. The frictional force provides a counterforce that opposes the movement of the block, performing work against the motion.

When friction is involved, the energy conservation equation changes to consider the work done by friction:

• Work done by friction: \( W_{friction} = -f \cdot L \)

Including this, our energy equation becomes \( mgh - W_{friction} = \frac{1}{2}mv_f^2 \). This equation underlines how energy lost to friction reduces the block’s final kinetic energy and consequently its speed.

The ability to factor friction into energy calculations is critical in real-world scenarios, where friction is almost always present.

Trigonometry in Physics

Trigonometry comes into play when we need to relate the height of the ramp to its length and angle, as done in this exercise. It is a mathematical tool that allows us to calculate angles and lengths, crucial in physics for analyzing problems involving inclines and slopes.

Here, we employ the sine function to find the angle of the ramp:

• \( h = L \sin\theta \)

Using this relation, we solve for the angle \( \theta \) by taking the inverse sine of the calculated ratio \( \sin\theta = \frac{h}{L} \).

Mastering trigonometry enables you to resolve components of forces, analyze projectile motion, and explore circular paths, thus unlocking a deeper understanding of physics in its many applications.