Emech at 1 + W = Emech at 2 mgh1 = mgh2 1. How high will pendulum rise? A) Less than h B) h C) More than h h From Tanner/Dubson CU Boulder Assuming no energy lost to anything else B Emech at 1 + W = Emech at 2 mgh1 = mgh2 Reference level (h = 0) קובץ זה נועד אך ורק לשימושם האישי של מורי הפיזיקה ולהוראה בכיתותיהם. אין לעשות שימוש כלשהו בקובץ זה לכל מטרה אחרת ובכלל זה שימוש מסחרי; פרסום באתר אחר (למעט אתר בית הספר בו מלמד המורה); העמדה לרשות הציבור או הפצה בדרך אחרת כלשהי של קובץ זה או כל חלק ממנו. Pendulum height

2. A 5000 kg coaster is released 20 meters above the ground on a frictionless track. What is the approximate speed at ground level? (point A) 7 m/s 10 m/s 14 m/s 20 m/s none of the above   Adapted from Franklin/Beale CU Boulder From rest D KEf+PEf=KE0+PE0 so 1/2 m v2+0=0+mgh v=sqrt(2gh)=20m/s Velocity from PE

3. What is its approximate speed at 10 meters high (point B )? A)       7 m/s B)       10 m/s C)       14 m/s D)      20 m/s E)        none of the above Adapted from Franklin/Beale CU Boulder C)       14 m/s KEf+PEf=KE0+PE0 so 1/2 m v2+mghf=0+mgh0 v=sqrt(2g(h0-hf))=14m/s  

4. How fast would the coaster have to be going at the start to reach 21 meters high (point C)? A)       1.1 m/s B)  3.2 m/s C)  4.5 m/s D)   20 m/s Adapted from Franklin/Beale CU Boulder C)       4.5 m/s  KEf+PEf=KE0+PE0 0+mghf=1/2 mv02 + mgh0 so v0=sqrt(2g(h-h0))=4.5 m/s There are more questions in the folder called KineticPotential

5. Calculate velocity at B if h = 0.45 m, g = 10 m/s2 A) 3 m/s B) 4.5 m/s C) 7.5 m/s D) 9 m/s A h From Tanner/Dubson CU Boulder still no energy lost A Reference level (h = 0) B Pendulum speed

CQ1 PE to KE from Dubson/Tanner 6. A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom.  To achieve a speed 2v at the bottom, how many times higher must the new ramp be? A)      B) 2     C) 3     D) 4    E) none of these. From Tanner/Dubson CU Boulder Answer: D. Using conservation of energy, we can write Ei = Ef.  If we place the zero of height at the bottom of the ramp, we have.. mghtop = 1/2 mvbottom2 So, the initial height h is proportional to the final speed-squared, v2 .  So to get twice the speed v, we need 4 times the height h.  [Since (2v)2 = 4v2]   v CQ1 PE to KE from Dubson/Tanner

A) B) 2 C) 3 D) 4 E) none of these. 6. A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom.  To achieve a speed 2v at the bottom, how many times higher must the new ramp be? A)      B) 2     C) 3     D) 4     E) none of these. mghtop + 0KE + 0work = 0PEg + 1/2 mvbottom2 First ramp: htop  vbottom2 2nd ramp: h'top  (2vbottom)2 = 4 (vbottom2) h'top = 4htop CQ 6 Show work