1. Examples: Mechanical Energy Conservation

2. Example 8.3 – Spring Loaded Cork Gun
Ball, mass m = 35 g = kg in popgun is shot straight up with spring of unknown constant k. Spring is compressed yA = m, below relaxed level, yB = 0. Ball gets to a max height yC = 20.0 m above relaxed end of spring.
(A) If no friction, find spring constant k.
(B) Find speed of ball at point B.
Ball starts from rest. Speeds up as spring pushes against it. As it leaves gun, gravity slows it down. System = ball, gun, Earth.
Conservative forces are acting, so use
Conservation of Mechanical Energy
Initial kinetic energy K = 0. Choose gravitational potential energy Ug = 0
where ball leaves gun. Also elastic potential energy Ue= 0 there. At max height, again have K = 0. Choose Note:
Need two types of potential energy!

3. Mechanical Energy is Conserved!! k = [2mg(yC – yA)/(yA)2] = 958 N/m
For entire trip of ball,
Mechanical Energy is Conserved!!
or: KA + UA = KB + UB = KC + UC.
At each point, U = Ug + Ue so,
KA+ UgA+ UeA = KB+ UgB + UeB = KC + UgC + UeC
(A) To find spring constant k, use:
KA+ UgA+ UeA = KC + UgC + UeC
or, 0 + mgyA + (½)k(yA)2 = 0 + mgyC + 0, giving
k = [2mg(yC – yA)/(yA)2] = 958 N/m
(B) To find ball’s speed at point B, use:
KA+ UgA+ UeA = KB + UgB + UeB
or, (½)m(vB) = 0 + mgyA + (½)k(yA)2 , giving
(vB)2 = [k(yA)2/m] + 2gyA; or, (vB) = 19.8 m/s

4. Example: Toy Dart Gun + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2
Mechanical energy conservation!
 (½)m(v1)2
+ (½)k(x1)2
= (½)m(v2)2
+ (½)k(x2)2
Speed when ?
leaves gun?
x1 = 0.06 m, v1 = 0, m = 0.1 kg, k = 250 N/m
x2 = 0, v2 = ? Find: v2 = 3 m/s ?

5. Example: Two Kinds of PE
m =2.6 kg
h =0.55 m
Y = 0.15 m
k = ?
A two step problem: STEP 1: (a)  (b)
 (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2
v1 = 0, y1 = h = 0.55 m, y2 =0 . Find: v2 = 3.28 m/s
STEP 2: (b)  (c) (both gravity & spring PE)
 (½)m(v2)2+(½)k(y2)2+mgy2 = (½)m(v3)2 + (½)k(y3)2 + mgy3
y3 = Y = 0.15m, y2 = 0  (½)m(v2)2 = (½)kY2 - mgY
Solve for k & get k = 1590 N/m
ALTERNATE SOLUTION: (a)  (c) skipping (b)
v1 = 0
v2 = ?
v3 = 0

6. Section 8.3: Problems with Friction
We had, in general:
WNC = K + U
WNC = Work done by non-conservative forces
K = Change in KE
U = Change in PE (conservative forces)
Friction is a non-conservative force! So, if friction is present, we have (WNC  Wf)
Wf = Work done by friction
In moving through a distance d, force of kinetics friction fk does work Wf = - fkd

7. When friction is present, we have:
Wf = -fkd = K + U = Kf – Ki + Uf – Ui
Also now, K + U  Constant!
Instead, Ki + Ui+ Wf = Kf + Uf
OR: Ki + Ui - fkd = Kf+ Uf
For gravitational PE:
(½)m(vi)2 + mgyi = (½)m(vf)2 + mgyf + fkd
For elastic or spring PE:
(½)m(vi)2 + (½)k(xi)2 = (½)m(vf)2 + (½)k(xf)2 + fkd

8. Example: Roller Coaster with Friction
m=1000 kg, d=400 m, y1=40 m, y2= 25 m, v1= y2 = 0, fk = ?
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + fkd
 fk= 370 N

9. Ex. 8.4 – Block Pulled on Rough Surface
A block, mass m = 6 kg, is pulled by constant horizontal force F = 12 N. over a rough horizontal surface. Kinetic friction coefficient μk = Moves a distance Δx = 3 m. Find the final speed.

10. Example 8.6 – Block – Spring System
A mass m = 1.6 kg, is attached to ideal spring of constant k = 1,000 N/m. Spring is compressed
x = cm = - 2  10-2 m & is released from rest.
(A) Find the speed at x = 0 if there is no friction.
(B) Find the speed at x = 0
if there is a constant friction force fk = 4 N.

11. Ex. 8.7 – Crate Sliding Down a Ramp
Crate, mass m = 3.0 kg, starts from rest at height yi = 0.5 m & slides down a ramp, length
d = 1.0 m & incline angle
θ = 30°. Constant friction force fk = 5 N. Continues to move on horizontal surface after.
(A) Find the speed at the bottom.
(B) Assuming the same friction force, find the distance on then horizontal surface that the crate moves after it leaves the ramp.

12. Ex. 8.8 – Block-Spring Collision
Block, mass m = 0.8 kg, gets initial velocity vA = 1.2 m/s to right. Collides with spring with constant k = 50 N/m.
(A) If no friction, find the maximum compression distance xmax of spring after collision.
(B) There is a constant friction force fk between block & surface. Coefficient of friction is μk = 0.5. Find the maximum compression distance xC now.

13. Ex. 8.9 – Connected Blocks in Motion
Two blocks, masses m1 & m2, are connected by spring of constant k. m1 moves on horizontal surface with friction. Released from rest when spring is relaxed when m2 at height h above floor. Eventually stops when m2 is on floor. Calculate the kinetic friction coefficient μk between m1 & table.