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When a 58 -g tennis ball is served, it accelerates from rest to a speed of 45 $\mathrm{m} / \mathrm{s}$ . The impact with the racket gives the ball a constant acceleration over a distance of 44 $\mathrm{cm} .$ What is the magnitude of the net force acting on the ball?

$130\mathrm{N}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

Lectures

04:16

In mathematics, a proof is…

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19:27

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to stop this question, we begin by discovering what is the acceleration off the ball for that we can use to Richard's equation, which tells us that the final velocity squared is acquitted. Initial velocity squared, plus two times the acceleration times the displacement. Let me trees my reference frame as being this X axis, pointing to the right. Then it goes as follows. The final velocity is 45 the initial velocity zero. They only have two times the acceleration times 44 centimeters. So we have to convert these two meters. These amounts to multiplying by 10 to the minus truth 44 centimeters is the same as 44 times. Stand to the miners truth meters. Then we have to solve this equation for the acceleration and it goes as follows. 45 squared is the course of two times 44 times 10 to the minus two times eight. Then a is equals to 45 squared, divided by true times 44 times down in the minors. Truth A is then equals to 45 squared, divided by 88 times down to the minors truth and then the acceleration is 45 squared, divided by 0.88 So now we can discover what is the magnitude off the net force that is acting on the bolt. For that, we have to use Newton's second law. The magnitude off the net forest acting on the boat is given by the mass off the ball times its acceleration. Then the mask off the boat is 58 grams share. You have to covertly two kilograms, and for that we multiply by 10 to the minus street, and then we multiply it by the acceleration, which is 45 squared, divided by 0.88 And these results in the net force off approximately 130 new tones. So this is the answer for this question.

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