Force Quiz
Chapter 5
Formulae
A.
Name: ____________________
Monday, October 26, 2009
QUIZ A
d = vi t + ½ a t2
vf2 = vi2 + 2 a d
F=ma
d = ½ (vf + vi) t
vf = vi + a t
Ff =  FN
g = 9.81 m/s2
FREE-BODY DIAGRAMS
2
Draw and label vectors a free-body diagram for:
a paper airplane while being thrown horizontally
Ff
a paper airplane after being released
finger/plane
Fair resistance
FA = Fnet
Fg
B.
Fg
DESCRIBING MOTION USING THE LAWS OF MOTION
3
Use your knowledge of the laws of motion to explain the following situation.
Describe each part of the motion and use each of the three laws of motion at least once.
Alex is doing push ups starting with his arms straight. He bends his elbows until his nose barely touches the floor.
Arms straight before bending his elbows Alex exerts a force downward on the floor equal to his weight.
The normal force of the floor exerts an equal and opposite force on Alex (3rd Law).
The force of gravity downward and the normal force upward produce no net force acting on Alex, so there is no
acceleration and Alex will continue to stay motionless (1st Law)
Elbows bending until his nose touches the floor When Alex bends his arms, he allows his weight to pull him down.
This means that his force down on the floor (and the floor up on him) is less than his weight down.
The net force is downward so he accelerates downward (2nd Law)
As his nose gets closer to the floor, he pushes harder than the gravitational force causing him to have an
upward acceleration as he goes down slower (2nd law)
C.
WORD PROBLEMS
7
g = 9.81 m/s2, [downward]
1. Find the acceleration of a 145 kg barbell when Aaron applies a force of 2350 N. [2 points] _____16.2 m/s2________
F=ma
a = F / m = 2350 N / 145 kg = 16.2 m/s2
2. Chris can throw a 140-gram baseball 85 miles per hour (32 m/s). The ball is at rest at the
start of his windup and it travels 2.5 m before it is released at a speed of 32 m/s.
What is the average force that Chris exerts on the baseball? [4 points]
______29 N_______
d = 2.5 m
vi = 0
vf = 32 m/s
a = 204.8 m/s2
t=
F = 29 N
m = 140 g = 0.140 kg
a = 204.8 m/s2
vf2 = vi2 + 2 a d
(32 m/s)2 = 0 + 2 a (2.5 m)
5.0 m
5.0 m
204.8 m/s2 = a
F = m a = (0.140 kg )(204.8 m/s2) = 28.672 kg m/s2 = 28.672 N
Force Quiz
Chapter 5
Name: ____________________
Monday, October 26, 2009
QUIZ B
Formulae
d = vi t + ½ a t2
vf2 = vi2 + 2 a d
F=ma
d = ½ (vf + vi) t
vf = vi + a t
Ff =  FN
g = 9.81 m/s2
A. WORD PROBLEMS
7
g = 9.81 m/s2, [downward]
1. In the muzzle of a rifle, a bullet accelerates from rest to a velocity of 500.0 m/s in 0.0020 s.
(a)
If the rifle produces a force of 9500 N, find the mass of the bullet. [3]
d=
vi = 0
vf = 500 m/s
a = 250 000 m/s2
t = 0.0020 s
F = 9500 N = 9500 kg m/s2
m = 0.038 kg
a = 250 000 m/s2
vf = vi + a  t
v f  v i (500m /s)  0
a

 250000m /s2
t
0.0020s

F=ma
F
9500N
m 
 0.038kg  38.0g
a 250000m /s2
2. Jesùs was going 12.5 m/s forward on his bike when he had to brake suddenly. His tires leave a long skid mark on the

dry pavement. The total mass of Jesùs and his bike is 75.5 kg
The kinetic coefficient of friction between his tires and the pavement is 0.80.
Find the acceleration of Jesùs [3]
___7.8 m/s2__________
Fnet Y = 0
Fg = m g = (75.5 kg) (9.81 m/s2) = 740.655 N [down]
Fnet x = Ff =  FN =  m g = (0.8) (740.655 N) = 592.524 N [backward]
a

______0.038 kg_______
Fnet 592.524N

 7.848m /s2 [backwards]
m
75.5kg
FN = 740.655 N [up]
B. DESCRIBING MOTION USING THE LAWS OF MOTION
3
Use your knowledge of the laws of motion to explain the following situation.
Describe each part of the motion and use each of the three laws of motion at least once.
Keisha throws a bowling ball down her lane and knocks down all of the pins.
Keisha swinging the ball in her hand
Keisha applies an upward force on the ball that is equal to its weight.
She also exerts an unbalanced forward force on the ball causing the ball to accelerate forward (2nd Law).
The ball on the alley after release
On the alley, the weight of the ball downward is balanced by the normal
force of the alley upward resulting in no net vertical force. There are no significant horizontal forces (friction is
small). There is no net force in any direction so there is no acceleration; the ball will roll at a constant speed in
a straight line (assuming no spin) (1st Law)
The normal force of the ball on the floor is equal to the normal force of the floor on the ball (3 rd Law)
1st Law
identify equal and opposite forces;
show that there is no net force and no acceleration; constant velocity
2nd Law
identify the net (unbalanced) force (include direction) causing the acceleration;
Fg
identify the forces affecting the net force and identify which one is greater
identify the change in velocity
3rd Law
C.
identify force pairs (FA on B = FB on A)
FREE-BODY DIAGRAMS
2
Draw and label vectors a free-body diagram for:
an object being pushed up an incline at a constant speed
FN
FA
Ff
Fg 
Fg
Fg ||
Parallel Forces
Fnet = FA + Ff + Fg ||
Fnet = 0, so FA = Ff + Fg ||
Perpendicular Forces
Fnet = FN + Fg
Fnet = 0, so FN = - Fg