Problem 19 - University Physics with Modern Physics

Question

Use the work-energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0 -m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 \(\mathrm{m}\) into the air. How fast was the boulder moving just as it left the volcano? (c) A skier moving at 5.00 \(\mathrm{m} / \mathrm{s}\) encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? (d) Suppose the rough patch in part (c) was only 2.90 m long? How fast would the skier be moving when she reached the end of the patch? (e) At the base of a frictionless icy hill that rises at \(25.0^{\circ}\) above the horizontal, a toboggan has a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) toward the hill. How high vertically above the base will it go before stopping?

Step-by-Step Solution

Answer

(a) 43.2 m/s, (b) 101.4 m/s, (c) 5.80 m, (d) 3.48 m/s, (e) 7.34 m.

Step 1: Understand the Work-Energy Theorem

The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it can be expressed as: \[ W = rac{1}{2} m v_f^2 - rac{1}{2} m v_i^2 \] where \( W \) is the work done, \( m \) is the mass, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity.

Step 2: Calculate the Velocity of the Falling Branch

For part (a), use the conservation of energy: the potential energy at the top is converted into kinetic energy at the bottom. The potential energy is \( mgh \), where \( g = 9.81 \, \text{m/s}^2 \) and \( h = 95.0 \, \text{m} \). Setting it equal to kinetic energy: \[ mgh = \frac{1}{2} mv^2 \] Simplify and solve for \( v \): \[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 95.0} \approx 43.2 \, \text{m/s} \] Thus, the branch's velocity is approximately 43.2 m/s when it reaches the ground.

Step 3: Determine the Initial Speed of the Ejected Boulder

For part (b), convert the kinetic energy at the starting point to potential energy at the highest point. The equation is: \[ \frac{1}{2} mv^2 = mgh \] Solving for \( v \) gives: \[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 525} \approx 101.4 \, \text{m/s} \] The boulder's speed as it leaves the volcano is approximately 101.4 m/s.

Step 4: Find the Distance the Skier Travels

For part (c), the work done by friction equals the initial kinetic energy. Using \( W_{friction} = f_k d = \frac{1}{2}mv_i^2 \), where \( f_k = \mu_k mg \): \( \mu_k = 0.220 \), \( v_i = 5.00 \, \text{m/s} \), and solving for \( d \): \[ \mu_k mgd = \frac{1}{2}mv_i^2 \] \[ d = \frac{v_i^2}{2\mu_k g} = \frac{5.00^2}{2 \times 0.220 \times 9.81} \approx 5.80 \, \text{m} \] The skier travels approximately 5.80 m before stopping.

Step 5: Calculate Skier's Speed after the Short Patch

For part (d), only part of the initial kinetic energy is dissipated by friction. Let \( v_f \) be the final speed after 2.90 m: \[ \frac{1}{2} m v_f^2 = \frac{1}{2} m v_i^2 - \mu_k mgd \] Solve for \( v_f \): \[ v_f = \sqrt{v_i^2 - 2\mu_k g d} = \sqrt{5.00^2 - 2 \times 0.220 \times 9.81 \times 2.90} \approx 3.48 \, \text{m/s} \] The skier's speed at the end of the patch is approximately 3.48 m/s.

Step 6: Compute the Height Toboggan Reaches

For part (e), convert initial kinetic energy to potential energy on the hill. \[ \frac{1}{2} mv_i^2 = mgh \] Solving for \( h \) gives: \[ h = \frac{v_i^2}{2g} = \frac{12.0^2}{2 \times 9.81} \approx 7.34 \, \text{m} \] Therefore, the toboggan reaches a height of approximately 7.34 m vertically.

Key Concepts

Kinetic Energy

Kinetic energy is all about the energy an object possesses due to its motion. When something moves, its kinetic energy comes into play. This can be described using the formula:

• \[ KE = \frac{1}{2} mv^2 \]
• where \( m \) represents mass and \( v \) represents velocity.

Imagine a ball rolling down a hill or a car cruising down the highway; the faster and heavier they are, the more kinetic energy they have. It's important to note that kinetic energy is always positive or zero since it depends on the square of velocity.
The conversion between different forms of energy often involves kinetic energy. For example, in free-fall situations like the falling branch from the tree, potential energy transforms into kinetic energy as the branch speeds up on its descent. This process is a practical demonstration of how kinetic energy increases as objects accelerate.

Potential Energy

Potential energy is the stored energy in an object due to its position or arrangement. A common type of potential energy is gravitational potential energy, which is affected by an object's height relative to a certain point, such as the ground.
A raised object, like the boulder ejected from the volcano, has gravitational potential energy calculated with:

• \[ PE = mgh \]
• Here \( m \) is mass, \( g \) is gravity (approximately 9.81 m/s²), and \( h \) is height.

When objects are placed at a height, they store energy which can be released by letting them drop, converting potential energy into kinetic energy. This is why potential energy is crucial in scenarios involving lifting or drop actions.
In exercises like the skier sliding down or a toboggan going up a hill, understanding potential energy is necessary to predict how high or how far objects can move when forces act upon them.

Friction

Friction is the force that opposes the motion of objects sliding against each other. It's a crucial concept that affects everything from cars on roads to skis on snow. The frictional force can be described simply by:

• \[ f_k = \mu_k N \]
• Where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force (usually \( mg \) for horizontal surfaces).

In the context of the skier in the problems, friction acts to slow down and eventually stop the skier.
When analyzing motion over rough patches of snow, the work done by friction (which is opposite to the motion) is vital to determine how far a skier can go before stopping. The laws of friction are essential for any situation involving surfaces that are not perfectly smooth. Friction converts some kinetic energy into thermal energy, contributing to energy loss from the system.

Conservation of Energy

The conservation of energy principle is fundamental in physics, stating that energy within an isolated system remains constant, though it can change forms. This means energy doesn’t just appear or disappear; it transforms from one type to another.
This principle is seen clearly in the conversion between potential and kinetic energy. For the exercises given, conservation of energy allows us to solve problems about falling objects or sliding on inclined planes without directly measuring forces or work.
Understanding that the sum of kinetic and potential energy at the beginning of a motion will equal the sum at another point, assuming no energy is lost, helps in predicting outcomes such as the final speed of fall or the height a sled reaches.

• For example, \[ \frac{1}{2} mv_i^2 + mgh_i = \frac{1}{2} mv_f^2 + mgh_f \]
• can be used to find unknown velocities or heights.

Even if energy might "appear" to be lost due to friction, it's actually converted into heat, preserving the energy balance in the broader system.