AP Physics – Newton’s Laws Test – 3 ans
1. (20) Batman throws a bat grenade at an angle of 44.0 to the horizontal with a speed of 17.8 m/s (the
Batman has a pretty good arm). How far does it travel?
m
m
m
m
vx  v cos   17.8 cos 44.00  12.80
v y  v sin   17.8 sin 44.00  12.36
s
s
s
s
m
m
12.36
 12.36
vy  vy 0
s
s
v y  v y 0  at t 

 2.522 s
m
a
9.8 2
s
x
m
vx 
x  vxt  12.80  2.522 s  
32.3 m
t
s
2. (20) What is the mass of a 335 N aardvark?



w
kg  m
1 
m
 335
34.2 kg
w  mg

 
g
s 2  9.8 m 

s 2 

3. (20) What net force is needed to accelerate a 2350 kg car at 4.80 m/s2?
m

F  ma  2350 kg  4.80 2   11 280 N 
11 300 N
s 

4. (20) When you jump into the air, you push the earth away from you and the earth pushes you away from
it. How come no one notices that you pushed the earth?
5. (20) A 2 350 kg gold nugget hangs from two cables which are at the angles shown.
Calculate the tensions in the two cables.
F
y
T
 0  T sin   T sin   mg  0

52
52
T

2T sin   mg
T
mg
2sin 
m

2350 kg  9.8 2 
s 


2  sin 52 
mg

14 600 N
6. (20) Two masses are connected by a light string which passes
over a frictionless pulley as shown. (a) What is the acceleration
of the system? (b) What are the tensions in the string?
m1a  T  m1g
m2a  m2 g  T
m1a  m2 a  T  m1 g  m2 g  T
m1a  m2a  m2 g  m1g
g  m2  m1 
 m1  m2 
T
0.545 kg
m1 g
 9.8
m2 g
0.515 kg
a  m1  m2   g  m1  m2 

m
0.277 2
 
s

m
m


m1a  T  m1 g T  m1 g  m1a  5.15 kg  9.8 2   5.15 kg  0.277 2  
s 
s 


a
T
m  5.45 kg  5.15 kg

s 2  5.45 kg  5.15 kg
51.9 N
7.
(20) You pull on a 98.0 kg box of bat pineapples and drag it across the deck. If the coefficient of kinetic
friction for the box and the deck is 0.405, what force must you apply to move the thing at a constant
speed?
m

F f 0
F  f   mg  0.405  98.0 kg   9.80 2  
389 N
s 

8. (20) A Find the acceleration of the system shown in the drawing if the coefficient of kinetic friction
between the 7.80 kg mass and the plane is 0.316.
m1a  T  m1g sin  m1g cos
m2a  m2 g  T
7.80 kg
6.90 kg
m1a  m2a  T  m1g sin  m1g cos  m2 g  T
m1a  m2a  m1g  m2 g sin  m1g cos
31.0
a  m1  m2   g  m2  m1 sin   m1 cos 
ag
a
 m2  m1 sin    m1 cos 
 m1  m2 
m
s2 
14.7
7.546
 9.8
m  6.90 kg  7.80 kg sin 31.0  0.316  7.80 kg  cos 31.0

s 2 
6.90 kg  7.80 kg
n
m
0.513 2
s
9. (20) Why do all objects fall at the same speed (ignoring air
resistance)?




T
T
f
FX
38.0
m2 g
m1 g
10. (20) The first law says that no force is required to maintain motion. Fine, then how come you have to
keep pedaling your bicycle to keep it moving?