### CH 5 PROBLEMSx

```253 m
A. Only the component of force in the
direction on the motion does work
W = Fd
= cos52(40)(253)
= 6230 J
B. If she is moving at constant speed,
the friction force must equal F|| and
the distance for friction is also 253 m,
so the WORK must also be the same
put in the opposite direction.
-6230 J
(I don’t get too worried if you don’t have the negative sign)
C. Use force diagram to see that:
FN +
F
= W
FN + sin52(40) = 70
FN = 38.5 N
Ff
= FN
cos52(40) =
 (38.5)
 = .640
A 50 km/hr
B
50km 1000 m 1 hr
--------x -----------x-----------= 13.9 m/s
hr
1km 3600 s
35 m
EA - Efriction = EB
½mv2
-
Ffd
=
½m(13.9)2 - (9.8m)(35)
½m(13.9)2
0
=
0
= (9.8m)(35)
.282 = 
Same situation, but with double the speed:
½m(27.8)2
27.8 m/s ……..find d
= (.282)(9.8m)d
140 m =
d
m= 20
W=196
sin25=h/d
25
EA
KE
-
½ m v2
-
½(20)(4)2
-
160
-
Efriction =
W
=
Ff d
FN
=
EB
PE
mgh
.2(cos25)(196)d = 20(9.8)(sin25)d
35.5d
=
d =
82.8d
1.35 m
A
This problem is just proving
that it doesn’t matter where
you draw the “zero line”
PEB = 0
10 m
B
Ref pt
A
A. PEA = 55(9.8)(10) = 5390
Ref pt
PE = PEA - PEB = 5390J
B. PEA = 0
PEB = 55(9.8)(-10)=-5390
10 m
B
PE = PEA - PEB = 5390J
C. PEA = 55(9.8)(5) = 2695
A
10 m
B
Ref pt
PEB = 55(9.8)(-5)=-2695
PE = PEA - PEB = 5390J
7.34
v?
EA
=
EB
PEgrav =
KE
mgh
= ½ mv2
9.8(7.34) =
.5v2
12.0 m/s =
v
37
A.
EA
=
EB
B.
EA
=
EB
70kg
4.0m/s
 = .70
EA
KE
2
½ mv
½ (70)(4)2
560
d?
-
Efriction =
EB
W
Ffd
=
=
0
0
=
0
- .70(70x9.8)d
-
480d
=
0
d = 1.17 m
ALL of the KE was lost due to friction, so the answer to part A is 560J
m = m
W = 9.8m
W|| = sin10.5(9.8m) = 1.79m
W = cos10.5(9.8m) =9.64m
EA
-
PEgrav
-
mgh
Efriction
Wf
-
Ffd
-
on hill
-
Wf
=
EB
=
0
on flat
Ffd
=
0
m(9.8)(36.4) - .075(9.64m)(200) - .075(9.8m)d
=
0
d
=
288 m
2 kg
k = 105 N/m
x = .10 m
EA
–
2 kg
.25 m
=?
Efriction
=
EB
Wfriction
Ff d
=
=
0
0
½ (105)(.1)2 - (29.8)(.25) =
0
PEspring
½ kx2
–
-
4.9 = .525
 = .107
```