1
Answer Section
HW 3
1. 4.1 m/s2, 52° north of east
2. 1.8 m/s2
3. 56 N
4. 12800 N
5. 4.8 kN, up
6. 150 N
7.
a. Fh – Ffr = 0
Fh = Ffr = 27 N
cos  = Fh / F = 27 N/43 N
 = 510
b. Fn + Fv – Fg = 0
(a = 0)
Fn = Fg - Fv = mg – F sin  = 18x9.8 – 43 sin 510
Fn = 140 N,
upward
c.k = Ffr /Fn = 27 N/ 140 N
k = 0.19
8. 0.75 m/s2
9. a) 1 m/s2
b) 10 N
10. 3.3 m/s2
11. 0.5
12. 20 kg
????
????
13. 290 N
14. A
T1 = T2 = T
2T sin 300 = 200
15. 100 N
16.
W = mg = (12.6 kg)(10.0 m/s2) = 126 N
2T sin = 126 N
T = 89 N
T=200N
2
17. 23N
m/s2
18. 4.90
5.1 m/s2
19. a=
x=41.6 m
20. 2.4 m/s2
21. 6.5 kg
Fn = mg cos = 532 N
Ffr =  Fn = 27 N
Fd = mg sin = 373N
Fnet = 346 N
d = ½ at2 = 12.9 m WOW
v = at = 11.7 m/s
22. W = mg = 650 N
Fnet = Fd – Ffr
a = Fnet/m = 5.32 m/s2
23.
Fn = mg cos = 8160 N
Ffr =  Fn = 408 N
Fd = mg sin = 3804N
Fnet = 4356 N
d = ½ at2
t = 6.43 s
v = at = 31.1 m/s
W = mg = 9000 N
Fnet = Fd – Ffr
a = Fnet/m = 4.84 m/s2
24. 0.69 m/s2, down
27. 8 N
GRAVITATIONAL FORCE
1.
E
2.
C
3. A
4.
If the moon orbited Earth at twice its present distance, the gravitational attraction between Earth and moon would be one
fourth as much. Because of this reduced attraction, the speed of the moon required to maintain a circular orbit would be less
(its present speed with less gravity would overshoot a circle, making an elliptical orbit). Because it would be moving more
slowly and also because it would have more distance to cover, the length of the month would increase.
5.
m:
my mass
M:
otherwise the Earth would orbit the Moon not the Sun
Earth mass
r:
Earth radius – distance from me and Earth
F = GmM/r2
Fnew = GmM/(0.25r)2 = 16 GmM/r2
= 16 F
the force is 16
times greater
6.
F = GmM/r2
Fnew = Gm(0.25)M/r2 =
¼ GmM/r2
=¼
F
the force is
4 times smaller
(1/ 4)
F = GmM/r2
7.
Fnew = GmM/(0.5r)2 =
4 GmM/r2
=4F
W = 4 x 70 =
8.
F = GmM/r2
Fnew = GmM/(5r)2 =
(1/25)GmM/r2
=
(1/25) F
9.
F = GmM/r2
Fnew = GmM/(3r)2 =
(1/9) GmM/r2
=
(1/9) W
10.
F = GmM/r2
Fnew = Gm(300)M/(11r)2 =
=
11.
(300/121) GmM/r2
(1/25) W
= (1/9)(72)(9.8) =
80 N
= (300/121) mg = (300/121)(2.5)(9.8)
61 N
g = GM/r2
gnew = GM/(1.1r)2 =
(1/1.21) GM/r2
= (1/1.21)g
280 N
= 8.1 N/kg = 8.1 m/s2
3
12.
g = GmM/r2
gnew = GM/(9 r)2 =
(1/81) GM/r2
=
(1/81)
g
=
0.1 N/kg =
0.1 m/s2
1
13.
F = GmM/r2
14.
F = Gm,m2/r2 = (6.67 X10-11 Nm2/kg2)(1 kg)(6 x1024 kg)I(6.4 X 106 rn)2 = 9.77 N (or rounded, 9.8 N)
15.
F = Gm1m2/(2r)2 = 9.8 N/4 = 2.44 N (or 2.45 N)
16.
F = Gm,m2/r2 =2.0x1020 N
17.
m1(2m2)/(2r)2 = ½
18.
W = m1m2/(2r)2 = ¼
19.
F = Gm,m2/r2
Fnew = Gm,m2/(1/10r)2 = 100 Gm,m2/r2 = 100 F
20.
F = Gm,m2/r2
Fnew = G(2m,)(2m2 )/(2r)2 =
21.
g = GMIr2 = 9.24 N/kg = 9.24 m/s2
22.
Fnew = GmM/(
3
r)2 =
9 GmM/r2
= 9 mg
9 times greater
as much
as much
G 4m,m2/4r2 =
9.24/9.8 = 0.94 or 94%
g = GM/r2 = 3 m/s2, so M gr2IG = 3(2 X 105)2 ÷ (6.67 x10-11 ) = 1.8 x 1021 kg
F
the force is 100 times greater
the force is unchanged