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Chapter 4: Problem 33

An \(80-\mathrm{kg}\) stuntman jumps from a window of a building situated \(30 \mathrm{~m}\) above a catching net. Assuming air resistance exerts a \(100-\mathrm{N}\) force on the stuntman as he falls, determine his velocity just before he hits the net.

Short Answer

Expert verified
The stuntman's velocity just before he hits the net is approximately \(23.1 \, m/s\).

Step by step solution

01

Determine the Gravitational Force

The force due to gravity can be calculated as the weight of the stuntman, which is his mass times gravity. Gravity is approximately \(9.8 \, m/s^2\), therefore the force due to gravity is \(F_g = m \cdot g = 80 \, kg \cdot 9.8 \, m/s^2 = 784 \, N\).
02

Calculate the Net Force on the Stuntman

The net force on the stuntman is the sum of the gravitational force and the air resistance (which acts in the opposite direction of the motion), so the net force is \( F_{net} = F_g - F_{airResistance} = 784N - 100N = 684N\).
03

Find the Acceleration of the Stuntman

The acceleration can be found using Newton's second law of motion, which states that the acceleration of an object is the net force on it divided by its mass. Therefore, acceleration \(a = F_{net}/m = 684N / 80kg = 8.55 m/s^2 \). Now, we must note that this is less than gravity due to the air resistance exerted on the stuntman as he falls.
04

Determine the Final Velocity

Finally, the velocity of the stuntman just before he hits the net can be calculated using the equation of motion: \(v^2 = u^2 + 2as\), where \(u\) is the initial velocity (which is 0, as the stuntman starts from rest), \(a\) is the acceleration, and \(s\) is the displacement. Substituting the values gives us: \(v^2 = 0 + 2 \cdot 8.55 \, m/s^2 \cdot 30m\), which simplifies to \(v = \sqrt{2 \cdot 8.55 \, m/s^2 \cdot 30m} = 23.1 \, m/s\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law of Motion
Newton's Second Law of Motion is a fundamental principle in physics that describes the connection between the motion of an object and the forces acting upon it. It states that the acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to its mass. The law is commonly written as the equation \( F_{net} = m \times a \), where \( F_{net} \) is the net force, \( m \) is the mass of the object, and \( a \) is the acceleration.

To better understand this concept, let's imagine pushing a shopping cart. If you apply a force to the shopping cart, it begins to accelerate in the direction of the force. If the same force is applied to a more massive cart, it accelerates less than the lighter one due to the inverse relationship with mass.
Gravitational Force Calculation
The gravitational force that an object experiences near the Earth's surface is what we commonly refer to as its weight. This force is calculated by the equation \( F_g = m \times g \), where \( F_g \) represents the gravitational force, \( m \) is the object's mass, and \( g \) is the acceleration due to Earth's gravity, which is approximately \(9.8 \, m/s^2\).

For our stuntman, with a mass of \(80 \, kg\), the gravitational force is calculated to be \(784 \, N\). It鈥檚 crucial to remember that while we often consider this force as constant, it does vary slightly depending on altitude and geographical location due to Earth's shape and density variations. This factor, however, is usually negligible for most practical problem-solving.
Final Velocity Determination
Determining the final velocity of an object in motion involves understanding its initial velocity, acceleration, and the distance over which the acceleration occurs. For an object starting from rest and moving under constant acceleration, the final velocity \(v\) can be found with the equation of motion \(v^2 = u^2 + 2as\), where \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.

In our example, the stuntman starts with an initial velocity of \(0 \, m/s\) and accelerates over a displacement of \(30 \, m\). Given the calculated acceleration, we determine his velocity just before landing in the net to be \(23.1 \, m/s\). This final velocity is an essential factor for various reasons, including safety calculations for stunts and understanding the motion of falling objects in physics.

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Most popular questions from this chapter

Two people are pulling a boat through the water as in Figure \(\mathrm{P} 4.24\). Each exerts a force of \(600 \mathrm{~N}\) directed at a \(30.0^{\circ}\) angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force \(\overrightarrow{\text { F exerted by the water on the boat. }}\)

The air exerts a forward force of \(10 \mathrm{~N}\) on the propeller of a \(0.20-\mathrm{kg}\) model airplane. If the plane accelerates forward at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\), what is the magnitude of the resistive force exerted by the air on the airplane?

An object of mass \(m\) is dropped from the roof of a building of height \(h .\) While the object is falling, a wind blowing parallel to the face of the building exerts a constant horizontal force \(F\) on the object. (a) How long does it take the object to strike the ground? Express the time \(t\) in terms of \(g\) and \(h\) (b) Find an expression in terms of \(m\) and \(F\) for the acceleration \(a_{x}\) of the object in the horizontal direction (taken as the positive \(x\) -direction). (c) How far is the object displaced horizontally before hitting the ground? Answer in terms of \(m, g, F_{t}\) and \(h\). \((d)\) Find the magnitude of the object's acceleration while it is falling, using the variables \(F, m\), and \(g\).

A dockworker loading crates on a ship finds that a \(20-\mathrm{kg}\) crate, initially at rest on a horizontal surface, requires a \(75-\mathrm{N}\) horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of \(60 \mathrm{~N}\) is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

A boat moves through the water with two forces acting on it. One is a \(2000-\mathrm{N}\) forward push by the water on the propeller, and the other is a \(1800-\mathrm{N}\) resistive force due to the water around the bow. (a) What is the acceleration of the \(1000-\mathrm{kg}\) boat? (b) If it starts from rest, how far will the boat move in \(10.0 \mathrm{~s}\) ? (c) What will its velocity be at the end of that time?
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