Work done when kicking a soccer ball

[alingy1]

Homework Statement

We want to calculate the work done by kicking the soccer ball. The acceleration of the ball is 6m/s^2. The mass of the ball is 0,5 kg. The distance the ball travels is 20 meters. The distance that the foot kicks the ball is 10 cm.

Homework Equations

W=Fd
F=ma

The Attempt at a Solution

My question is whether we need to use the distance of 20m or 10 cm. The concept of work is new to me. I don't understand if the distance in the formula W=Fd refers to the distance that the force travels while being applied to the soccer ball. Or does it refer to the distance the ball travels (20m). Can you please explain why? I really want to fully understand the idea of work. W=Fd is just too vague for me, especially since I want to continue my studies in physics. Thanks.

 

[Dick]

Homework Statement

We want to calculate the work done by kicking the soccer ball. The acceleration of the ball is 6m/s^2. The mass of the ball is 0,5 kg. The distance the ball travels is 20 meters. The distance that the foot kicks the ball is 10 cm.

Homework Equations

W=Fd
F=ma

The Attempt at a Solution

My question is whether we need to use the distance of 20m or 10 cm. The concept of work is new to me. I don't understand if the distance in the formula W=Fd refers to the distance that the force travels while being applied to the soccer ball. Or does it refer to the distance the ball travels (20m). Can you please explain why? I really want to fully understand the idea of work. W=Fd is just too vague for me, especially since I want to continue my studies in physics. Thanks.

The d in W=Fd is the distance over which the force is applied. 10 cm.

 

[CWatters]

Once the ball leaves the foot the foot can't do any more work on the ball. So it's 10cm.

 

[HallsofIvy]

If an object with initial speed 0 has acceleration "a" m/s^2 for t seconds, then its final speed will be at and it will have gone a distance (1/2)at^2. Knowing that a= 6 m/s^2 and that it has gone 10 cm= .1 m, you can calculate the time required by solving (1/2)(6)t^2= 3t^2= 0.1. Use that time to find its final speed and then its kinetic energy. The work done is the same as the increase in energy.

 

[Nickie]

Hello, thank you for your question. I would like to clarify that the work done when kicking a soccer ball refers to the energy transferred to the ball from the foot. In this case, we would use the formula W=Fd, where F is the force applied to the ball and d is the distance over which the force is applied.

In this scenario, the distance of 10 cm (0.1 m) would be used as the distance in the formula, as this is the distance over which the force of the kick is applied to the ball. The distance that the ball travels (20 m) is not relevant to the calculation of work done, as it is the result of the force applied to the ball, not the distance over which the force is applied.

I understand that the concept of work may be new to you, and it can be a bit abstract at first. In simple terms, work is the energy transferred to an object by a force acting on it. In this case, the force of the kick transfers energy to the ball, causing it to move. The amount of work done is directly proportional to the force applied and the distance over which it is applied.

I hope this explanation helps you understand the concept of work better. It is an important concept in physics and is used in many real-world applications. If you have any further questions, please do not hesitate to ask. Good luck with your studies in physics.