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A football punter accelerates a football from rest to a speed of \(10 \mathrm{~m} / \mathrm{s}\) during the time in which his toe is in contact with the ball (about \(0.20 \mathrm{~s}\) ). If the football has a mass of \(0.50 \mathrm{~kg}\), what average force does the punter exert on the ball?

Short Answer

Expert verified
The average force the punter needs to exert on the ball is 25 N.

Step by step solution

01

Calculate the acceleration

We start by calculating the acceleration of the ball. Acceleration is defined as the change in velocity over time. Since the initial velocity of the ball is 0 (as it was at rest), the change in velocity is equal to the final velocity. So, acceleration \(a\) is calculated as \(a = \frac{\Delta v}{t}\), where \(\Delta v = 10 m/s\) (final velocity) and \(t = 0.20 s\). Thus, \(a = \frac{10 m/s}{0.20 s} = 50 m/s^2\).
02

Calculate the force

Next, we calculate the force using the formula \(F = ma\), where \(m = 0.50 kg\) (mass of the ball) and \(a = 50 m/s^2\) (acceleration calculated previously). Therefore, \(F = 0.50 kg * 50 m/s^2 = 25 N\)

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Key Concepts

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

Acceleration Calculation
When we talk about acceleration, we're referring to the rate of change in velocity that an object experiences. It's an indication of how quickly an object speeds up or slows down. In physics, acceleration is commonly denoted by the symbol `a` and is calculated using the formula \[\begin{equation} a = \frac{\text{{change in velocity}}}{\text{{time}}}\end{equation}\]
Regarding our football example, we start with the ball at rest which means it has an initial velocity of 0 m/s. When the ball reaches a speed of 10 m/s, all within 0.20 seconds of contact with the punter's foot, we can use the change in velocity (\[\begin{equation}\Delta v\end{equation}\]= 10 m/s) and the time (t = 0.20 s) to calculate the acceleration using the formula \[\begin{equation} a = \frac{\Delta v}{t}\end{equation}\]
.Simple arithmetic gives us \[\begin{equation}a = \frac{10 m/s}{0.20 s} = 50 m/s^2\end{equation}\]
, showing that the ball's acceleration is 50 meters per second squared. This calculation gives us insight into the rapid velocity change the ball undergoes due to the punter's action.
Force Calculation
To understand how a football can be kicked so far, we must look into the force exerted on the ball. Force, represented by the symbol `F`, is essentially a push or a pull upon an object resulting from its interaction with another object. Calculating force requires knowing two important pieces of information: the object's mass and its acceleration. According to the second law of motion put forth by Sir Isaac Newton, the force can be found using the equation \[\begin{equation} F = ma\end{equation}\]
. In our football scenario, we've already discovered the ball's acceleration is 50 m/s^2. The ball's mass, given in the problem, is 0.50 kilograms. Plugging these values into the force equation gives us \[\begin{equation} F = 0.50 kg * 50 m/s^2\end{equation}\]
. Upon calculating, this results in a force of \[\begin{equation} F = 25 N\end{equation}\]
(Newtons). This tells us that the punter is applying a force of 25 Newtons to the football to achieve the given acceleration.
Newton's second law
Digging deeper, Newton's second law of motion is a cornerstone of classical mechanics and is crucial for explaining how and why objects move the way they do. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. Mathematically, it is expressed as `F = ma`.Newton's second law connects the concepts of force, mass, and acceleration in a single elegant equation, allowing us to predict the resultant force when the other two quantities are known. The beauty of this law lies in its universal applicability; it holds true for a wide range of situations, from the motion of planets to the kick of a football. Recognizing the influence of mass and acceleration helps us understand that the same force will cause a lighter object to accelerate more than a heavier one. Correspondingly, in our exercise, the relatively low mass of the football allows it to accelerate swiftly when a significant force is applied by the punter. Understanding Newton's second law empowers students to solve various physics problems by linking these fundamental concepts.

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Most popular questions from this chapter

A magician pulls a tablecloth from under a \(200-\mathrm{g} \mathrm{mug}\) located \(30.0 \mathrm{~cm}\) from the edge of the cloth. The cloth exerts a friction force of \(0.100 \mathrm{~N}\) on the mug and is pulled with a constant acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\). How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than \(30 \mathrm{~cm}\) relative to the tabletop during the process.

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of \(80.0 \mathrm{~N}\) at an angle of \(25.0^{\circ}\) above the horizontal. The box has a mass of \(25.0 \mathrm{~kg}\), and the coefficient of kinetic friction between box and floor is \(0.300 .\) (a) Find the acceleration of the box. (b) The student now starts moving the box up a \(10.0^{\circ}\) incline, keeping her \(80.0 \mathrm{~N}\) force directed at \(25.0^{\circ}\) above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?

A chinook salmon has a maximum underwater speed of \(3.0 \mathrm{~m} / \mathrm{s}\), and can jump out of the water vertically with a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). A record salmon has a length of \(1.5 \mathrm{~m}\) and a mass of \(61 \mathrm{~kg}\). When swimming upward at constant speed, and neglecting buoyancy, the fish experiences three forces: an upward force \(F\) exerted by the tail fin, the downward drag force of the water, and the downward force of gravity. As the fish leaves the surface of the water, however, it experiences a net upward force causing it to accelerate from \(3.0 \mathrm{~m} / \mathrm{s}\) to \(6.0 \mathrm{~m} / \mathrm{s}\). Assuming the drag force disappears as soon as the head of the fish breaks the surface and \(F\) is exerted until two-thirds of the fish's length has left the water, determine the magnitude of \(F\).

A shopper in a supermarket pushes a loaded cart with a horizontal force of \(10 \mathrm{~N}\). The cart has a mass of \(30 \mathrm{~kg}\). (a) How far will it move in \(3.0\) s, starting from rest? (Ignore friction.) (b) How far will it move in \(3.0 \mathrm{~s}\) if the shopper places his \(30-\mathrm{N}\) child in the cart before he begins to push it?

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.

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