Download Dynamics Problems in 2D Worksheet Answer Key

Physics 12
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Dynamics Problems in 2D Worksheet Answer Key
1. A 25.0 kg sled is accelerating at 2.3 m/s2. A force of 300.0 N is pulling the sled along
a rope that is being held at an angle of 35o with the horizontal. What is the
coefficient of friction?
m 25.0kg
Fp 300.0N
 35
?
Fg mg
(25.0)(9.80)
245N
a x 2.3m / s 2
We must first calculate the components for the applied force:
Fpx Fp cos
(300.0) cos 35
246N
Vertical:
Using up as positive,
ma y Fy
may FN Fg Fpy
0 FN Fg Fpy
FN Fg Fpy
245 172
73N
Horizontal:
Fpy Fp sin
(300.0) sin 35
172N
max F x
max Fpx Ff
max Fpx Ff
(25.0)(2.3) 246 Ff
Ff 189N
Ff FN
189 (73)
 2.6
2. A 15.0 kg sled is being pulled along a horizontal surface by a rope that is held at a
20.0o angle with the horizontal. The tension in the rope is 110.0 N. If the
coefficient of friction is 0.30, what is the acceleration of the sled?
m 15.0kg
Fp 110.0N
 20.0
0.30
ax ?
Fg mg
(15.0)(9.80)
147N
We must first calculate the components for the applied force:
Fpx Fp cos
(110.0) cos 20.0
103N
Fpy Fp sin
(110.0) sin 20.0
37.6N
Vertical:
Using up as positive,
ma y Fy
may FN Fg Fpy
0 FN Fg Fpy
FN Fg Fpy
147 37.6
109N
Horizontal:
Ff FN
(0.30)(109)
33N
max F x
max Fpx Ff
max Fpx Ff
(15.0)ax 103 33
ax  4.7m / s 2
3. A man pushes a 15 kg lawnmower at constant speed with a force of 90.0 N
directed downward along the handle, which is at an angle of 30.0o to the
horizontal. What is the coefficient of friction?
m 15kg
Fp 90.0N

 30.0
?
a x 0
a y 0
Fg mg
(15)(9.80)
147N
We must first calculate the components for the applied force:
Fpx Fp cos
(90.0) cos 30.0
77.9N
Fpy Fp sin
(90.0) sin 30.0
45.0N
Vertical:
Using up as positive,
ma y Fy
may FN Fg Fpy
0 FN Fg Fpy
FN Fg Fpy
147 45.0
192N
Horizontal:
ma x F x
max Fpx Ff
max Fpx Ff
0 77.9 Ff
Ff 77.9N
Ff FN
77.9 (192)
 0.406
4. A sled is being pulled by a rope that makes an angle of 27o with the
horizontal. The coefficient of friction between the sled and the ground is 0.30.
If the tension in the rope is 245 N and the acceleration of the sled is 1.2 m/s2,
what is the mass of the sled?
m ?
Fp 245N
 27
 0.30
a x 1.2m / s 2
a y 0
We must first calculate the components for the applied force:
Fpx Fp cos
(245) cos 27
220N
Vertical:
Using up as positive,
may Fy
may FN Fg Fpy
0 FN Fg Fpy
FN Fg Fpy
9.80m 110
Horizontal:
Ff FN
(0.30)(9.80m 110)
2.9m 33
Fpy Fp sin
(245) sin 27
110N
max F x
max Fpx Ff
max Fpx Ff
m(1.2) 220 (2.9m 33)
1.2m 220 2.9m 33
4.1m 253
m  62kg
5. A 55.0 kg rock is being pulled along a horizontal surface at a constant speed.
The coefficient of friction between the rock and the surface is 0.76. If the rope
pulling the rock is at a 40.0o angle with the horizontal, with what force is the
rock being pulled?
m 55.0kg
Fp ?
 40.0
 0.76
a x 0
a y 0
Fg mg
(55.0)(9.80)
539N
We must first calculate the components for the applied force:
Fpx Fp cos
Fpy Fp sin
Fp cos 40.0
Fp sin 40.0
0.766Fp
0.643Fp
Vertical:
Using up as positive,
may Fy
may FN Fg Fpy
0 FN Fg Fpy
FN Fg Fpy
539 0.643Fp
Horizontal:
Ff FN
0.76(539 0.643Fp )
410 0.49Fp
max F x
max Fpx Ff
max Fpx Ff
0 0.766Fp (410. 0.49Fp )
0 0.766Fp 410. 0.49Fp
410 1.26Fp
Fp  330N
6. A 40.0 kg iceboat is gliding across a frozen lake with a constant velocity of 14
m/s E, when a gust of wind from the southwest exerts a constant force of 100.
N on its sails for 3.0 s. With what velocity will the boat be moving after the
wind has subsided? Ignore any frictional forces.
m 40.0kg
vi 14m / s E
t 3.0s
F 100. N NE
v f ?
a 
Fnet ma
100. (40.0)a
2.50 
v
t
v
3.0
v 7.5m / s NE
a 2.50m / s 2 NE
v v f vi
v f vi v
vfx
vf
vfy
vy
v
vi
vx
vx vy 7.5cos 455.3m / s
v fx vi vx
14 5.3
19m / s
v f  v 2fx v 2fy
 (19)2 (5.3)2
20.m / s
v fy vy
5.3m / s
tan 
v fx
v fy
19
5.3
 74
v f 20. m / s, 74E of N
7. Two tow trucks attach ropes to a stranded vehicle. The first tow truck pulls
with a force of 25000 N, while the second truck pulls with a force of 15000 N.
The two ropes make an angle of 15.5o with each other. Find the resultant
force on the vehicle.
This problem can be simplified by aligning one of the forces along the x-axis
F1 25000N
F2 15000N
Fnet ?
Fnet F1 F2
F2 y F2 sin15.5
F2x F2 cos15.5
(15000) cos15.5
14000N
(15000) sin15.5
4000N
Fnetx F1 F2x
Fnety F2 y
25000 14000
39000N
F  (F
net
) (F
2
netx
4000N
)
2
F
tan  Fnety
netx
nety
 (39000)2 (4000)2
39000N
Fnet 39000N,5.9from the larger force
4000
39000
 5.9