Ap mechanics problem

Test Preparation AP Test Preparation
1

<p>Hey guys I was doing previous years frq and got stuck in one solution. </p>

<p>Question 3 part b-
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

<p>Solution by collegeboard-
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

<p>In the solution they have taken work done = change in KE but the skier is also going up the hill therefore his potential energy is increasing too. How can we neglect this??
Also in the alternate solution hoe do we substitue for dx in terms of dt??</p>

2

<p>Oh just realised the dx and dt substitution part. Though potential energy part is still unclear.
Can you help me out please?</p>

3

<p>I believe that’s just the definition of the work-energy theorem. The work done is just the change in kinetic energy.</p>

<p>Also think about this. The change in kinetic energy is also equal to the change in potential energy. So, in a way, you’re not really neglecting PE.</p>

<p>Not sure if that helped but good luck!</p>

4

<p>The question asks what is the work done by the NET force. The net force includes both the rope pulling the skier and the force of gravity, so the force of gravity and gravitational potential energy are accounted for in the work.</p>

<p>edit: If they asked what is the work done by the rope, it would be equal to the change in KE plus the change in PE.</p>

5

<p>“The change in kinetic energy is also equal to the change in potential energy. So, in a way, you’re not really neglecting PE.” </p>

<p>This is true if the energy of the skier is constant. However, here the rope is doing work on the skier, increasing his energy, so it is not true that the change in KE = change in PE.</p>

6

<p>I didn’t read the question carefully, I was too quick to assume. Thanks for the correction utche11!</p>

7

<p>Sorry for taking over the post, but can anyone help me with the 2003 FRQ Question 2 Part C? </p>

<p>The part that says Mg=kD, shouldn’t you add the mass of the pan AND clay, ending up with 2Mg=kD?? Is this a typo on the scoring guideline?</p>

<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

8

<p>@jdroidxw They have taken mg =kd before the clay impacts the tray. Because after it impacts it will cause an increase in the extension of the spring. Since we already know from before the collision a value of k there is no need to calculate net extension of the spring due to 2m</p>