Chapter 10 : Do listed problems--8, 13, 14, 16, 21, 22, 23, 24, 31, 38, 41,  49, 55

Some problems could deal with earlier chapters, which is why I assigned them i.e. to give you an integrated view. You should find many of them fun if you get past the time it takes to solve them. However, this quiz should not be to time consuming.

Please note this list includes online problems which can be done for EC after you enroll in the course at http://www.masteringphysics.com/ . Read RECENT  email.
Simulations
Quiz 10
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8. From conservation of energy the height should be 1.0 m. Use ½mv12 + mgy1 = ½mv22 + mgy2.  Find y2. 
13. Use ½mv12 + mgy1 = ½mv22 + mgy2.  Find v2. 
14. Fx = -kx  also Fx = max. So find ax,  given k and x. Convert x to meters. 
16. 
(a) max = pos - neg 
0 = kd - mg, where d is the distance the spring stretches from un-deformed  length. You must subtract the two lengths  given to get d. We see that kd = mg. Find k for m = 2.0 kg.  

(b) Then find the new distance stretched when m = 3.0 kg. 
21. ½mv12 + ½kx12 = ½mv22 + ½kx22. Here, and the problems below,  x is the distance the spring is stretched or compressed from equilibrium defined as x = 0. Note:  v1 = 0. Also x2 = 0. Find v2   if  x1 = -0.04 m. 
22. 
Again, ½mv12 + ½kx12 = ½mv22 + ½kx22  .   Note:  v1 = v and v2  = 0 .  Also x1 = 0 and x2 = - 0.02 m.   Find v .
After you find v, re-do problem to find the new x2  if  v1 = 2v  and  x1 = 0 and v2  = 0. The new x2  should be twice the old  x2 in magnitude. 
23. ½mv12 + ½kx12 = ½mv22 + ½kx22  . Note the final velocity and x1 are  0. 
24. ½mv12 + ½kx12 = ½mv22 + ½kx22  . Note the final velocity and x1  are 0. 
31. This is a cool problem, briefly discussed in principle  when we started Chapter 14 in lab and referenced spring potential energy: 
U = ½kx2. The curve of Figure Ex10.31 looks somewhat like ½kx2  except  it's raised, shifted  and asymmetric. 
Use ½mv12 + U1 = ½mv22 + U2   .  Now,  the maximum speed occurs when U is a minimum (at x = 4.0 mm).  The speed is zero when x = 2.0 mm or 8.0 mm.  Let the first point be x1 = 2.0 mm. Let the second point x2 = 4.0 mm.   Use the above conservation of energy equation to find |v2| = speed at x2. Note  there are two solutions for v2 since the particle could be moving in the positive or negative x-directions. We  want the absolute value, the speed. 
38.  In (b), I would assume a larger  speed than (a)  when the block leaves the spring (at x = 0).  Why? Because in (b) you have twice the potential energy when the spring and mass are released from rest. 
Here's a hint: 
(a) ½mv12 + ½kx12 = ½mv22 + ½kx22   ,  where the initial velocity (at 1)  and second position (at 2) are zero.  Here v2 = vo. 
(b) ½mv12 + ½(2k)x12 = ½mv22 + ½kx22   ,  where the initial velocity (at 1)  and second position (at 2) are zero.  Here v2 must be solved for in terms of the vo in (a).  

In both cases, v2 is written in terms of |delta x| . Solve for them in each case, and compare.  Note that x12 = |delta x|2 in each case. 
41.  ½mv12 + mgy1 +  ½kx12 = ½mv22 + mgy2 + ½kx22   .  Now |x1| = 0.10 m. Set y1 = 0 when the spring is compressed. v1 = 0,  so all you have is spring potential energy spring at 1. Also, |x2| = 0, since the package has already left the spring when it arrives at the top (2). In addition,   v2 = 0.  Solve for y2. Note y2 is the vertical displacement of the package from the bottom (1).  To get the distance traveled along the 30 degree incline, divide by sin30.