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Chapter 4: Problem 2

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 exerted by the punter on the football is \(25 \, \mathrm{N}\).

Step by step solution

01

Calculate Acceleration

Acceleration can be calculated using the formula \( a = \frac{{v - u}}{{t}} \), where \( v \) is the final speed, \( u \) is the initial speed, and \( t \) is time. Given that \( v = 10 \, \mathrm{m/s} \), \( u = 0 \, \mathrm{m/s} \) (since the ball was at rest), and \( t = 0.20 \, \mathrm{s} \), we can plug these into the formula to get the acceleration: \( a = \frac{{10 - 0}}{{0.20}} = 50 \, \mathrm{m/s^2} \).
02

Apply Newton's Second Law

According to Newton鈥檚 second law of motion, the force exerted on an object is equal to the mass of the object multiplied by its acceleration, which can be stated as \( F = m \cdot a \). In this case, mass \( m = 0.50 \, \mathrm{kg} \) and acceleration \( a = 50 \, \mathrm{m/s^2} \).
03

Calculate Average Force

Using the given values for mass and acceleration, the force can be calculated as \( F = 0.50 \, \mathrm{kg} \times 50 \, \mathrm{m/s^2} = 25 \, \mathrm{N} \). So, the average force that the punter exerts on the football is 25 Newtons.

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Most popular questions from this chapter

After falling from rest from a height of \(30 \mathrm{~m}\), a \(0.50-\mathrm{kg}\) ball rebounds upward, reaching a height of \(20 \mathrm{~m}\). If the contact between ball and ground lasted \(2.0 \mathrm{~ms}\), what average force was exerted on the ball?

The force exerted by the wind on the sails of a sailboat is \(390 \mathrm{~N}\) north. The water exerts a force of \(180 \mathrm{~N}\) east. If the boat (including its crew) has a mass of \(270 \mathrm{~kg}\), what are the magnitude and direction of its acceleration?

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 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?

(a) An elevator of mass \(m\) moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, \(T\) or \(w ?\) (b) When the elevator is moving at a constant velocity upward, which is greater, \(T\) or \(w\) ? (c) When the elevator is moving upward, but the acceleration is downward, which is greater, \(T\) or \(w ?\) (d) Let the elevator have a mass of \(1500 \mathrm{~kg}\) and an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of \(10 \mathrm{~m} / \mathrm{s}\). Find T. Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with your answer to part (c)?
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