Question

A girl on her sled has just slid down a hill onto a level field of ice and is starting to slow down. At the instant when their speed is \(5 \mathrm{ft} / \mathrm{sec}\), the girl's father runs up and begins to push the sled forward, exerting a constant force of \(15 \mathrm{lb}\) in the direction of motion. The combined weight of the girl and the sled is \(96 \mathrm{lb}\), the air resistance (in pounds) is numerically equal to one-half the velocity (in feet per second), and the coefficient of friction of the runners on the ice is \(0.05\). How fast is the sled moving \(10 \mathrm{sec}\) after the father begins pushing?

Short answer

The sled is moving at approximately \(10.112 \,\mathrm{ft/s}\) 10 seconds after the father begins pushing.

Step-by-step solution

Identify the forces acting on the sled

There are three forces acting on the sled: 1. The force exerted by the father, which is a constant force of 15 lb. 2. The air resistance force, which is numerically equal to half of the velocity (in feet per second). 3. The friction force between the sled and the ice, with a coefficient of friction of 0.05. We can compute the normal force as the combined weight of the girl and the sled (\(96\text{ lb}\)).

Find the friction force on the sled

The friction force (\(F_{friction}\)) can be calculated using the formula: \( F_{friction} = μ \cdot F_{normal} \) Where μ is the coefficient of friction (0.05) and \(F_{normal}\) is the normal force (96 lb). So, \( F_{friction} = 0.05 \times 96 = 4.8 \,\mathrm{lb} \)

Set up the differential equation using Newton's second law of motion

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (∑F = ma). After considering the friction force and air resistance, the net force acting on the sled is: \( 15 - 4.8 - \frac{1}{2} v(t) = m a(t) \) The mass of the girl and sled can be found by dividing their weight by the acceleration due to gravity (\(32 \,\mathrm{ft/s^2}\)): \( m = \frac{96}{32} = 3 \,\mathrm{lb}\cdot \mathrm{s^2/ft} \) We also know that acceleration is the derivative of velocity, so: \(a(t) = \frac{dv}{dt}\) Substitute m and a(t) in the equation: \( 15 - 4.8 - \frac{1}{2}v(t) = 3\frac{dv}{dt} \)

Solve the differential equation for the velocity function

First, let's separate variables: \( \frac{dv}{dt} = \frac{10.2 - \frac{1}{2}v(t)}{3} \) Multiply both sides by dt and integrate: \( \int_{0}^{T} \frac{2}{3} dt = \int_{5}^{v(T)} \frac{1}{10.2 - \frac{1}{2}v(T)} dv \) Now, solve the integrals: \( \frac{2}{3}t \Big|_0^T = -2 \ln\left| \frac{1}{2}v(T) - 10.2 \right| \Big|_5^{v(T)} \)

Calculate the velocity of the sled 10 seconds after the father begins pushing

Now, substitute T with 10 and solve for the velocity: \( \frac{2}{3} \cdot 10 = -2 \cdot [ \ln\left|\frac{1}{2}v(10)- 10.2 \right| - \ln\left|\frac{1}{2}(5)- 10.2 \right| ] \) Solve for \(v(10)\): \( v(10) \approx 10.112 \,\mathrm{ft/s} \) Thus, the sled is moving at approximately 10.112 ft/s 10 seconds after the father begins pushing.

Key concepts

Newton's Second Law of Motion

Understanding Newton's Second Law of Motion is crucial for solving problems related to forces and motion. In essence, it states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This relationship can be formulated as \( F_{net} = m \times a \), where \( F_{net} \) represents the net force, \( m \) is the mass, and \( a \) is the acceleration.

When we apply this principle to a real-world scenario like a sled being pushed on ice, we must consider all the forces acting on the sled: the applied force, friction, and air resistance. By summing up these forces and setting this equal to the mass times acceleration, a differential equation is obtained that can model the sled's motion over time. Solving this equation provides us with the sled's changing velocity, which directly ties into the second core concept: velocity.

Velocity

Velocity is a vector quantity that describes both the speed and direction of an object's motion. In the context of our sled problem, velocity tells us how fast the sled is moving in a specific direction—in this case, along the ice field. As forces act on the sled, the velocity changes; this rate of change of velocity is known as acceleration.

Mathematically, we can express velocity as the first derivative of position with respect to time \( v(t) = \frac{dx}{dt} \) and acceleration as the derivative of velocity \( a(t) = \frac{dv}{dt} \). Understanding these relationships allows us to predict the future speed of the sled at any given time, which is exactly what we do when we solve the ordinary differential equation derived from Newton's second law.

Friction Force

Friction force is a resistive force that acts opposite to the direction of motion whenever two surfaces come in contact. This force is what makes the sled slow down as it slides across the ice. The magnitude of friction force (\( F_{friction} \) can be calculated using the equation \( F_{friction} = \mu \times F_{normal} \), where \( \mu \) is the coefficient of friction, and \( F_{normal} \) is the normal force—essentially, the force pressing the two surfaces together, which for our problem, is the weight of the girl and the sled.

The lower the coefficient of friction, the less the friction force, and the smoother the slide. With ice being quite slippery, the coefficient is low, resulting in a relatively minor force compared to rougher surfaces, but it is still significant enough to influence the sled's motion.

Air Resistance

Air resistance, often referred to as drag, is the force that opposes an object's motion through air. It increases with the velocity of the object and can be affected by the shape, size, and speed of the moving object. In the sled example, the problem states that the air resistance is numerically equal to one-half of the velocity (in pounds) \( F_{air} = \frac{1}{2}v(t) \).

This creates a dynamic situation where, as the sled accelerates or decelerates, the air resistance changes in proportion to its speed. It is a critical component that must be incorporated into the net force calculation when applying Newton's second law. Ignoring air resistance would lead to an incorrect prediction of the sled’s motion and an unrealistic understanding of the physics involved.