<p>a mass of 12kg is placed on an incline angled at 30 degrees from the horizontal. a spring that can be compressed 2 cm by a force of 270N is placed on the incline below the mass. the mass is released and hit the spring and momentarily stops when it compressed the spring by 5.5 cm. how far has the mass moved down the incline? why is the speed of the mass when it hits the spring?</p>
<p>the correct answer for the first question is 35cm
answer for second question is 1.7m/s</p>
<p>I worked it out but got the wrong answer. my calulation is like below:</p>
<p>Energy of spring = kx^2 = 270N/2cm * 5.5cm * 0.055m = 40.83J
Energy of spring = kinetic energy = (m * v^2) / 2.
solve for v i got 2.608 (the correct answer is 1.7)</p>
<p>Kinetic energy = the work done on the mass by the component of gravity
= F * d = 9.8 * 12 * cos60 * d
solve for d I got 0.694m (correct answer 0.35m)</p>
<p>wut did i do wrongly??</p>
<p>plz help me</p>
<p>For the first part, I solved for the weight of the 12 kg mass along the incline's plane, and then used that to solve for the distance it traveled before hitting the spring...</p>
<p>Wx = mg sin 30° = 12(9.80) sin 30° = 58.8 N</p>
<p>Now we can use the work equation to start to get somewhere...
Work = Fd = 58.8d</p>
<p>When the mass hits the spring, it is transferring all of its energy from the work shown above to potential energy of the spring, so</p>
<p>0.5(13500)(0.0550)^2 = 58.8d
0.35 m = 35 cm = d</p>
<p>For the second part, you could solve it in a variety of ways, but I keep getting 1.8 m/s and not 1.7 m/s. Here's what I did anyway...</p>
<p>Using kinematic equations, Vf = (Vo^2 + 2ad)^(1/2), and in this scenario, Vo = 0, a = g sin 30° = 4.90 m/s^2, and d = 0.35 m. Plugging in reveals that Vf = 1.8 m/s.</p>
<p>Using energy equations, all kinetic at the tip of the spring equals all potential when it is compressed 0.055 m, so
0.5(13500)(0.055)^2 = 0.5(12)v^2
1.8 m/s = v</p>
<p>Hope this helps, even if the speed is off by 0.1 m/s :).</p>
<p>Happy New Year all! :D</p>
<p>0.1 m/s is fine, I think i remeber the formula wrong.
Thank you very much, Happy new year:)</p>