Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. As usual, we'll start the question by writing the stuff down that we know. The shotput is 7 kilograms and travels through a distance of 2.8 meters, its final speed is 13 m/s, and we assume its initial speed is zero. We also have to assume that the force is exerted just horizontally so that we don't have to consider gravity in our free-body diagram or calculations. We'd need more information like the angle and so on if we were to have to account for gravity. So, we assume that it's horizontal. And from this picture then we can see that the force on the shotput is the only force horizontally and so it is the net force. And because it's the net force we can use mass times acceleration to calculate it. And so, we need to find the acceleration then. And this formula here has everything we know except for acceleration. And we know the initial speed is zero. And we'll you know get rid of that and switch the sides around and then divide both sides by 2d, giving us acceleration is the final velocity 30 meters per second, we square that and divide by 2 times the distance 2.8 meters. And that gives us acceleration 30.1786 m/s squared which we then multiplied by mass, 7 kilograms, to get the force applied on the shotput with two significant figures of 210 Newtons.