March 26, 2009.

Hi!

Today, we corrected the chapter 10 study guide and the problem 1-6 on page 330 ( duck book)

Answers (chapter 10 study guide)
Work:
force, distance, direction, w= Fd, work, force, distance, scalar, joule, newton, meter, joule,moves, force displacement.

Work and direction of force:
in the direction of, perpendicular, in the direction of, cosine,direction of motion, negative, energy, energy, motion,positive,negative.

Power:
rate, rate, energy, P=W/t, power, work, time, watt, joule,one second, kilowatt.

Answer (problem 1-6 on page 330 duck book)
1a. w= Fd (d=15cm. F= 40N)
= 40N x 0.15m
=6 N/m or 6 J.
1b. W= Fd (m= 50 kg, d=1.95m, g=9.8)
=mgd
=50kg x 9.8N/kg x 1.95m=956 J
1c. W= Fdcos25o
=120N x 4m x cos25o
=435J
2. W= Fd (a=2.5m/S2, V2= 12.5m/s, t=5s, F=5000N)

V1= V2 - at=12.5m/s - ( 2.5 m/S2 x 5s)
= 0

d=( V1+V2)/2 x t
= (0+12.5m/s)/2 x 5s
=31.25m

W= Fd
= 5000N x 31.25m
= 156250J or 1.6 x 10^5 J

3. (F= 78N , m = 10m)
angle = 90 - 35 = 550

W= Fdcos550
= 78N x 10m x cos550
= 447.4 J

4.( V1= 25 m/s, V2= 14m/s, t= 5s, m= 52000kg)

d= (V1+V2)/2 x t
= (25 + 14)/2 x 5= 97.5m

a= (V2- V1)/t
= (14 - 25)/5
= -2.2 m/S2

F=ma
= 52000 x (-2.2m/S2)
= -114400N

W= Fd
= 114400 x 97.5
= 1154000J

5a. w= Fd
= 175 x 55
= 9625 J

5b. W= A1 + A2 + A3
= (xy/2) + (xy/2) + xy
= [ (0.04 x (-20))/2] + [ (0.04 x 20)/2] + [(0.12-0.08) x 20]
= (-0.4J) + (0.4J) + (0.8J)
= 0.8J

6. ( W= 480 J, m=3 kg, g= 9.8)
W= mgd

d= W/ mg
= 480J/ 3 x 9.8
=16.3m


Today, Ms. K gave us some worksheet to work onChapter 11 study guideWorksheets on chapter 10 and 11 study guide.

The next scribe is Abbas.
Bye!