1. 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)
Pendulum height

2. 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. 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. 4. How fast would the coaster have to be going at the start to reach 21 meters high (point C)?
A)       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. 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

6. 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

7. 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