Chapter 8Conservation of Energy EXAMPLES

Example 8.1 Free Fall(Example 8.1 Text book) Determine the speed of the ball at y above the

ground The sum K + U remains constant At h: Ui = mgh Ki = 0 At y: Kf = ½mvf

2 Uf = mgy In general: Conservation of Energy

Ki + Ui = Kf + Uf

0 + mgh = ½mvf2 + mgy

Solving for vf

vf is independent of the mass !!!

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Example 8.2 Roller Coaster Speed

Using Mechanical Energy Conservation (Frictionless!) ½mvf

2 + mgyf = ½mvi2 + mgyi

Only vertical differences matter! Horizontal distance doesn’t matter! Mass cancels!

Find Speed at bottom?Known: yi = y = 40m, vi = 0,

yf = 0, vf = ?

0 + mgyi = ½mvf2 + 0

vf2 = 2gyi = 784m2/s2

v2 = 28 m/s

Example 8.3 Spring-Loaded Gun (Example 8.3 Text book)

Choose point A as the initial point and C as the final point

(A). Find the Spring Constant k ?

Known: vA = 0, yA = 0 , xA = yB = 0.120m vC = 0 , yC = 20m, UC = mgyC , m = 35.0g

EC = EA KC + UgC + UsC = KA + UgA + UsA

½mvC2 + mgyC + ½kxC

2 = ½mvA2 + mgyA + ½kxA

2

0 + mgyC + 0 = 0 + 0 + ½kxA2

½kxA2 = mgyC k = 2mgyc/xA

2

=2(0.0350kg)(9.80m/s2)(20.0m)/(0.120m)2

k = 953 N/m

yA

yB

yC

Example 8.3 Spring-Loaded Gun, final

(B). Find vB ?

Use: EB = EA

KB + UgB + USB = KA + UgA + USA ½mvB

2 + mgyB + ½kxB2 =

½mvA2 + mgyA + ½kxA

2 ½mvB

2 + mgyB + 0 = 0 + 0 + ½kxA2

vB2 = (kxA

2 – 2mgyB)/m vB

2 =388.1m2/s2

vB = 19.7 m/s

yA

yB

yC

Example 8.4 Ramp with Friction(Example 8.7 Text book)

Problem: the 3.0 kg crate slides down the rough ramp. If: vi = 0, yi = 0.5m , yf = 0

ƒk = 5N (A). Find speed at bottom vf

At the top: Ei = Ki + Ugi = 0 + mgyi

At the bottom: Ef = Kf + Ugf = ½ m vf2 + 0

Recall: If friction acts within an isolated system ΔEmech = ΔK + ΔU = Ef – Ei = – ƒk d

½ m vf2 – mgyi = – ƒk d

Solve for vf

smvsmdfmgym

v fkif /54.2/47.62 222

Example 8.4 Ramp with Friction, final

(B). How far does the crate slide on the horizontal floor if it continues to experience the same friction force ƒk = 5N

The total ΔEmech is only kinetic (the potential energy of the system remains fixed):

ΔEmech = ΔK = Kf – Ki = – ƒk d Where: Kf = ½ m vf

2 = 0

Ki = ½ m vi2 = 9.68 J

Then: Kf – Ki = 0 – 9.68 J = – (5N)d

d = (9.68/5) = 1.94 m

Example 8.5 Motion on a Curve Track (Frictionless)

A child of mass m = 20 kg starts sliding from rest. Frictionless!

Find speed v at the bottom. ΔEmech = ΔK + ΔU

ΔEmech =(Kf – Ki) + (Uf – Ui) = 0

(½mvf2 – 0) + (0 – mgh)= 0

½mvf2 – mgh = 0

Same result as the child is falling vertically trough a distance h!

smghv f /3.6)2)(10(22

Example 8.5 Motion on a Curve Track (Friction)

If a kinetic friction of ƒk = 2N acts on the child and the length of the curve track is 50 m, find speed v at the bottom.

If friction acts within an isolated system ΔEmech = ΔK + ΔU = – ƒk d

ΔEmech = (Kf – Ki) + (Uf – Ui) = – ƒk d

ΔEmech = ½mvf2 – mgh = – ƒk d

ΔEmech = ½(20) vf2 – 20(10)(2) = –100J

smvsmdfmgym

v fkif /5.5/302 222

Example 8.6 Spring-Mass Collision(Example 8.8 Text book)

Frictionless! K +Us = Emech remains constant

(A). Assuming: m= 0.80kg vA = 1.2m/s k = 50N/m

Find maximum compression of the spring after collision (xmax)

EC = EA KC + UsC = KA + UsA

½mvC2 + ½kxmax

2 = ½mvA2 + ½kxA

2

0 + ½kxmax2 = ½mvA

2 + 0

max0.80 (1.2 / ) 0.1550 /A

m kgx v m s mk N m

Example 8.6 Spring-Mass Collision, final (B). If friction is present, the energy decreases by:

ΔEmech = –ƒkd Assuming: k= 0.50 m= 0.80kg vA = 1.2m/s k = 50N/m

Find maximum compression of the spring after collision xC

ΔEmech = –ƒk xC = –knxC = –kmgxC ΔEmech = –3.92xC (1) Using: ΔEmech = Ef – Ei

ΔEmech = (Kf – Uf) + (Ki – Ui)ΔEmech = 0 – ½kxC

2 + ½mvA2 + 0

ΔEmech = – 25xC2 + 0.576 (2)

Taking: (1) = (2): – 25xC2 + 0.576 = –3.92xC

Solving the quadratic equation for xC :xC = 0.092m < 0.15m (frictionless) Expected! Since friction retards the motion of the system

xC = – 0.25m Does not apply since the mass must be to the right of the origin.

Example 8.7 Connected Blocks in Motion(Example 8.9 Text book)

The system consists of the two blocks, the spring, and Earth. Gravitational and potential energies are involved

System is released from rest when spring is unstretched.

Mass m2 falls a distance h before coming to rest.

The kinetic energy is zero if our initial and final configurations are at rest K = 0

Find k

Example 8.7 Connected Blocks in Motion, final

Block 2 undergoes a change in gravitational potential energy

The spring undergoes a change in elastic potential energy

Emech = K + Ug + US = Ug + US = Ugf – Ugf + Usf – Usi

Emech = 0 – m2gh + ½kh2 – 0

Emech = – m2gh + ½kh2 (1)

If friction is present, the energy decreases by: ΔEmech = –ƒkh = – km1gh (2)

Taking (1) = (2): – m2gh + ½kh2 = – k m1gh

k m1gh = m2gh – ½kh2

gmkhgm

k1

21

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