_______Initials
Name: _____Answer Key_______
Date: ____________________________
Physics Test

Major Grade
Form A
FORMULA BANK:
Press. = Force
Area

a= v
t
FNET = FX - Ff
FW = mg, weight
Ff = FN
For a block on an inclined plane, FX = Fwsin
For a block on an inclined plane, FY = Fwcos
On a horizontal surface, FW = FN
FX = Ff in a simple equilibrium situation on a
flat surface or inclined plane
On an inclined plane,
FY = FN
FNET = ma
Distance traveled: x = ½at
2
MATHEMATICAL PART.
DIRECTIONS:
Where applicable, show your work in the space provided for partial credit and put your
answer in the blank provided for it. The points for each problem are shown next to the problem.
Over-Easy
1. (five points) Draw and label the real forces
acting on the box in the Figure at right.
Friction is present. Assume that equilibrium
exists.
FN3
box
The real forces DON’T involve breaking down diagonal forces
into the “cajita feliz”
2. (five points) Draw and label all the
real forces acting on Block A on the
inclined plane in the Figure at right.
Assume the blocks want to slide down
the plane. Friction is present.
FN1
FN2
Fw
Ff
FN FT1
FT2
A
Ff
Fw
Since the blocks want to slide down the plane,
friction goes up, always opposite the motion.
Medium-Rare
3. A box of mass 5 kg. is being dragged along the floor at a
constant velocity of 10 m/s by a force of 30 N.
____0.61___ (one point) What is the coefficient of friction,?
The force of friction, Ff = Fx, the horizontal pushing force, when the object is moving at constant velocity.
There is NO acceleration so the forces to the left and right cancel out. So if Fx = 30 N, then Ff = 30 N. We
also know that on a horizontal surface, the Normal Force, FN = FW, the weight.
We know that the weight, FW = mg, so also, FN = mg.
Putting all the formulas together, we can rewrite the formula for friction Ff = FN as Fx =  mg.
Plugging in our values and solving for , the coefficient of friction:
30 N = (5 kg)(9.8 m/s2) → 30 N = 49→ 30 N = 49 → 0.61= 
49
49
_______Initials
_6929.24 N_4. (one point) If the mass of the sign is 630 kg, what is the tension in
the rope?
If the mass of the sign is 630 kg, then its weight is
given by FW = mg = 630 kg  9.8 m/s2 = 6174 N.
We know that in an equilibrium situation, Fy = Fw.
So we can also say that Fy = 6174 N.
Fy
FT
63º
FT= ? N
Dr. Bones
Fy=6174 N
Chiropractor
63°
To find FT, we need to use S.O.H.C.A.H.T.O.A. Since I know the opposite
side and am trying to find the hypotenuse, I will use S.O.H.
opp.
Fy
6174 N
→sin 63 =
sin  =
→sin  =
→ FT  sin 63 = 6174 N
hyp.
FT
FT
FT  sin 63 = 6174 N → FT = 6929.24 N
sin 63 sin 63
Fw
_111.8, 63.44°_5. (three points) Find the exact resultant of the forces shown
at right, magnitude and direction.
1. Line up the forces head-to-tail.
2. Draw the resultant vector from the tail of the
first to the head of the last.
3. Count how many squares form the sides of the
right triangle formed by the resultant vector.
4. Find the resultant force using the Pythagorean
Theorem: a2 + b2 = c2
12 + 22 = c2 → 1 + 4 = c2 → 5 = c2
√5 = √c2 →2.24 squares = c.
2 squares
5.
Use the scale given to find your resultant force. If 5
squares are 250 N, then each square = 50 N so 2.24
squares = 2.24 x 50 N = 111.80 N
1 square
6. To find the direction, find the horizontal angle using
S.O.H.C.A.H.T.O.A. You can use any trig function
that you want at this point because you have all sides – opposite, adjacent, and hypotenuse.
250 N
sin  = opp./hyp. → sin  = 2/2.24 →  = sin (2/2.24) →  = 63.44°
-1
cos  = adj./hyp. → cos  = 1/2.24 →  = cos (1/2.24) →  = 63.44°
-1
tan  = opp./adj. → tan  = 2/1 →  = tan (2) →  = 63.44°
-1
_______Initials
Well-Done
Fx
6. (two points) A 2-kg block slides down the ramp
shown in 2.5 seconds when friction is not present.
55º
_25.09 m_ How long is the ramp?
The only formula that has distance in it (how long) is
Fx = opp.
x = ½at2.
I know the time, t, is 2.5 seconds but I need the
acceleration, a, in order to find x, the distance.
55º
How do I find “a” when there is no friction?
First I need to realize that the acceleration of the
Fy = adj.
block down the inclined plane is because a force, Fx,
was applied to it (see at right):
Fw = hyp.
nd
Fx = ma, according to Newton’s 2 Law
According to my formula bank, Fx = Fwsin  (see the right triangle shown above)
So if Fx= ma and Fx = Fwsin, then
ma = Fwsin (substituting…set them equal to each other)
But we know that the formula for weight, Fw = mg, so
ma = mg sin (substituting for Fw)
the mass cancels out of the equation, leaving…
ma = mg sin
m
m
a = gsin.
Plugging in our values: a = 9.8 m/s2 sin(55) = 8.03 m/s2. We now know acceleration.
Plugging “a” into our distance equation along with the time, “t”:
x = ½at2= ½(8.03 m/s2)(2.5 s)2 = 25.09 m
OVER-cooked
(five points) Essay. This essay will
be graded with the 6+1 Writing
Traits. Write your answer in the
A
B
spaces provided. Assume that Block A
16º
has a mass of 0.6 kg Block B has a
mass of 1.8 kg. and that there is no
friction on the slope Block B is on but there is friction on the slope Block A is on (coefficient
of friction of  = 0.8). Explain:
(a.) …what forces to the left and what forces to the right would have to be balanced for an
equilibrium situation to exist. HINT: there are two forces. Name them correctly.
The only force to the left is FfA, the force of Friction
between Block A and the surface it is resting on. The
FfA
only force to the right is FxB, the horizontal component
of the weight, Fw, of Block B.
A
B
16º
FxB
_______Initials
(b.) …how equilibrium could possibly exist if Block B is so much more massive than
Block A.
Although Block B is much bigger than Block A, there is friction on the slope that Block
A is on, and this works opposite to the direction of motion. In addition, the angle of
the incline that Block B is on is very little (16°), so it doesn’t want to slide that
much.
(c.) …what formula(s) you would use to find the force(s) pulling to the left (that you stated
in Probl. A.)
The force to the left is the friction force, FfA. The formula for the force of friction,
FfA = FNA. On a flat horizontal surface we know that, FNA = FwA, and we know that
the weight, FwA = mAg. Substituting we get: FfA = mAg
(d.) …what formula(s) you would use to find the force(s) pulling to the right (that you stated
in Probl. A.)
The force to the right is the horizontal component of the weight of Block B, FxB. On an
inclined plane in an equilibrium situation, Fx = FwBsin B, and we know that the weight,
FwB = mBg. Substituting we get: FxB = mBg sin B
(e.) …whether or not the system above is actually in
equilibrium, by finding the force(s) pulling to the left
and comparing them to the force(s) pulling to the right.
For the system to be in equilibrium, the force(s) to the left
would have to equal the force(s) to the right. Substituting:
Forcesleft = Forcesright
FfA = FxB
6 + 1 Grading Rubric
________ /5 (65%) Ideas & Content
________ /5 (5%) Organization
________ /5 (15%) Word Choice
________ /5 (5%) Voice
mAg = mBg sin B
In this case do not be tempted to cross out the masses.
________ /5 (5%) Sentence Fluency
They are different from each other and cannot be cancelled.
________ /5 (5%) Conventions
mAg = mBg sin B
Substitute your given values to see if the system is in equilibrium:
_________ /5 OVERALL GRADE
0.8(0.6 kg)9.8 m/s2 = 1.8 kg 9.8 m/s2  sin (16°)
4.704 N  4.86 N
The force to the right is not the same as the force to the left so
there is a resultant force of Fresultant = 4.86 – 4.704 = 0.156 N to the right. The system will slide
that way.
CONCEPTUAL PART.
DIRECTIONS: Put the letter of the correct answer in the blank. Each question is worth 1 point.
____c.____
If an object is falling through the air, the force of air resistance will be pointing
a. to the right
d. downwards
b. to the left
e. not enough information
c. upwards
Friction, including air resistance, always acts opposite the direction of motion. If an object is falling
DOWN, air resistance will point UPWARDS.
_______Initials
__b.__
You are moving a box along a plane. Which is more efficient: pulling a load
with a rope at an angle  of 50° or at an angle  of 30° with the horizontal?
a. at the 50° angle.
c. the same for both.
b. at the 30° angle.
d. the angle is irrelevant.
The lower the angle , the greater the horizontal
component of the force, Fx. Fx is what moves the box
horizontally. See at right.
Fp

Fp


Fx
Fx
___a._
For an object lying on an inclined plane, the normal force, FN, on the object by the plane is equal and
opposite to which force?
a. Fy
b. Fx
c. FT
d. Fw
e. Ff
FN
On an inclined plane, FN = Fy. They cancel each other out. On a flat horizontal surface,
FN = Fw
FY
__a.__
If the Force of Friction gets bigger, what does that mean about the Normal Force?
a. It gets bigger.
d. It gets bigger and then smaller.
b. It gets smaller.
e. It gets smaller and then bigger.
c. the two are not related to each other.
The friction force, Ff, is proportional to the Normal Force, FN. The relationship is given by Ff = FN. Whatever
friction does (get bigger or smaller), the normal force will do too.
__b.__
An object is stationary on a horizontal surface. Friction is present. The object at first experiences no
horizontal force. A slowly increasing force is then applied to it in the horizontal direction. Which describes
best what will happen?
a. the object will move immediately at a constant speed. Friction disappears instantly.
b. the object will not move until the force applied equals the maximum force of starting friction and
then it will move.
c. the object will move slowly until the force applied equals the maximum force of starting friction.
d. the object will move immediately at an accelerating speed.
Friction opposes motion until a certain point. It will exactly equal Fx or Fp until a maximum point where it cannot
resist anymore. After this breakaway point, the object being pushed will move at constant velocity. If more Fp or Fx
is applied, then the object will actually accelerate.
F
f
An object lies still on an inclined surface that has an angle  of 55°. Fx will be
Fx
a. the same as Fy.
d. the same as Ff.

b. less than FN.
e. the same as Fw.
c. less than Fy.
If the object lies still on the inclined plane, it is in equilibrium. If it is in equilibrium, the forces up the plane and
the forces down the plane must be equal and opposite. Therefore, Fx = Ff.
__d.__
___d.__
How many "real" forces are acting on the sign post? Assume it has mass
a. one
d. four
b. two
e. five
c. three
FT1
FT2
FN
FW
Fy
Fy FT
All other things being equal, which picture hanging from the


ceiling will break easier of the two pictures shown at right?
a. the picture to the left
b. the picture to the right
c. the forces on the strings are the same
FW
FW
d. the angle the strings make don’t affect the forces
In order for the picture to hang and not fall, the force upwards (Fy) has to equal the force downwards (FW). The
force to do this comes from the diagonal strings holding up the pictures. Because the angle with the horizontal is
smaller in the picture at right, the Force of tension, FT, required to produce Fy is much greater. To lift the same
weight, Fw, the strings holding up the picture on the right have to be tighter and are thus more likely to break.
___b.__
FT