Physics - Unit V Review
**Considering the frictionless situation at left, understand how acceleration might
change if one or more of the features change (if mass changes, or force changes) by
filling in increases, decreases, stays the same, not enough info to tell. When
F
answering, consider Newton’s 2nd Law solved for acceleration  a =
m
1. If the mass of the cart is cut in half, the acceleration Increases. (same force, but half the mass)
2. If the towing force is decreased to 0.25N, the acceleration Decreases. (half the force, but same mass)
3. If the mass and force is decreased by half, the acceleration Same. (if both decrease by half, acc is the same)
4. If the force is applied at an angle above the horizontal, the acceleration Decreases. The same force is at an
angle, but not in the direction of acceleration, so there is only a component of the 0.5N force (smaller than 0.5N)
in the direction of acceleration (Smaller force, same mass)
5. If the track is inclined
the acceleration Decreases. Now there is a horizontal
component in the opposite direction of the 0.5N force, so the sum of the forces is 0.5N minus the component .
(Smaller force, same mass)
6. If the 0.5 N force is applied to 4 carts (each 0.25kg) all connected together, the acceleration Stays the same. 4
carts added together = 1kg. (Same mass same force)
**Given a velocity vs time graph, draw the corresponding a vs t and x vs t graphs. (Unit 3 Stacks of Kinematic
Curves, or Unit 3 Kinematic Curves Quiz)
7. For the v vs t graph at right, between section A or C,
where is the net force the smallest & why?
V
C, the acceleration at C is the smallest which means
a smaller net force.
B
C
A
8. For the graph at right, is there a section
where net force equals zero? Why?
B and D, the acceleration is zero during at those points
which means a net force of zero.
©Modeling Workshop Project 2006
D
1
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Unit V review v3.0
From dynamics information - If you are given forces, or the physical description of the system and
surroundings, draw a force diagram. Ask yourself: "Can I tell if the system is accelerating?" If yes, then the
forces do NOT add up to zero. ΣF = ma. If the system is moving at constant velocity or is motionless, then the
forces cancel out. ΣF = 0.
System #1
System #2
9. For system #1: The coefficient of kinetic friction is 0.1, and the masses are 15kg and 2kg, find the acceleration of
the system.
−𝟏𝟗.𝟔𝑵+𝟏𝟒.𝟕𝑵
Fk = µFn = µ(Fg) = 0.1(15kg)(9.8m/s2) = 14.7N
𝒂 = 𝟏𝟓𝒌𝒈+𝟐𝒌𝒈 = −𝟎. 𝟐𝟗 𝒎⁄ 𝟐 since it is considered to be
𝒔
one object (2kg mass and 15kg mass connected together) then we ignore the negative sign. There is a pull force
causing acceleration (2kg mass = 19.6N) and a 14.7N force opposing the pull.
10. For system #2: Mass is 10kg, and the frictional force is 25.36N. Draw a rotated force diagram where the normal
force is pointing up in the positive Y-axis and the frictional force is pointing in the positive x-direction. What is
the value of μ?
Fgx = Fgsinθ = 10kg(9.8m/s2)sin15 = 25.36N
Fgy = Fgcosθ = 10kg(9.8m/s2)cos15 = 94.66N
Fgx =Frictional force Fs (not accelerating in x-direction)
Fgy = Fn (not accelerating in y-direction)
𝑭
𝟐𝟓.𝟑𝟔𝑵
Fs = μFn  𝝁 = 𝑭 𝒔 = 𝟗𝟒.𝟔𝟔𝑵 = 𝟎. 𝟐𝟕
𝑵
11. a) For system #3: mass is 20kg, pull force is 200N, what is the acceleration? (Consider frictionless since µ = 0)
𝟐𝟎𝟎𝑵𝒄𝒐𝒔𝟐𝟎
sum forces in direction of acceleration  -0N + Fpx = ma  𝒂 = 𝟐𝟎𝒌𝒈 = 𝟗. 𝟒m/s2
only the component in the x direction is in the direction of acceleration
b) Consider system #3 again, but now μk = 0.1, what is the acceleration?
To get the frictional force, we need to determine the normal force
-Fg+Fpy+FN = 0  FN =Fg – Fpy = 196N – 200Nsin20 = 127.6N
Fk = μkFN = 0.1(127.6N) = 1.276N
To get the acceleration, sum the forces in the x-direction
−𝟏.𝟐𝟕𝟔+𝟐𝟎𝟎𝑵𝒄𝒐𝒔𝟐𝟎
-Fk+Fpx = ma
𝒂=
= 9.3m/s2
𝟐𝟎𝒌𝒈
©Modeling Workshop Project 2006
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Unit V ws4 v3.0
12. Given kinematic information (∆x, v, t), find the acceleration first, then use F = ma to solve for force.
a) A 12,000 kg bus slows from 30 m/s to 10 m/s in 10 s. What is the net force acting on the bus?
𝟏𝟎𝒎 𝟑𝟎𝒎
−
Need to find acceleration first 𝒂 = 𝒔 𝟏𝟎𝒔 𝒔 = -2m/s2
Now we can find the net force  F = ma = 12000kg(-2m/s2) = -24000N
b) What does a scale read for a 75 kg man in an elevator that goes from - 6.0 m/s to zero in 2.0 s ?
𝟎𝒎
𝟔𝒎
−(− )
𝒔
𝒔
Need to find acceleration first  𝒂 =
= 3m/s2 This positive acceleration means the person is moving
𝟐𝒔
downward (-6m/s) and is slowing down (accelerating up while moving down slows an object down).
Force diagram
-Fg + FN = ma
FN = Fg + ma
FN = 75kg(9.8m/s2) + 75kg(3m/s2) = 960N
13. A 10 kg box is dragged across a horizontal surface ( = 0.20) by a 100 N force that is applied at a 20° angle
from the horizontal. Describe how you would go about finding the acceleration of the box. (You can solve the
problem mathematically, and then describe step by step, and why you took that step)
Steps
1.
Step 1.
2.
3.
Step 2.
4.
20o
Fpy = 100Nsin20 = 34.2N
Fpx = 100Ncos20 = 94.0N
5.
6.
Fg = 10kg(9.8m/s2) = 98N
I would first draw a picture of the situation to visualize what forces
are on the object.
Next, I would draw a force diagram and fill in as much information
as you can before going further.
After obtaining all the information on the force diagram, the next
thing is to find the normal force (can’t get the acceleration without
the frictional force, and the frictional force depends on the normal
force).
So to find the normal force, sum the forces in the direction of the
normal force (y-direction) and solve for normal force.
a. –Fg + FN + Fpy = 0
b. FN = Fg - Fpy = 98N - 34.2N = 63.8N
Now find frictional force using Fk = µFN = 0.2(63.8N) = 12.8N
Finally, I can sum the forces in the direction of acceleration (the x
−12.8𝑁+100𝑁𝑐𝑜𝑠20
direction) –Fk + Fpx = ma  𝑎 =
= 𝟖. 𝟏 𝒎⁄ 𝟐
10𝑘𝑔
𝒔
14. An 80 kg water skier is being pulled by a boat with a force of 220 N which causes the skier to accelerate at
1.8 m/s2. Find the drag force on the skier.
Sum the forces in the direction of acceleration (x-direction) & solve for FD
-FD + FP = ma  FD = Fp – ma = 220N – 80kg(1.8m/s2) = 76N
©Modeling Workshop Project 2006
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Unit V ws4 v3.0
15. A 2000 kg car is slowed down uniformly from 20 m/s to 5 m/s in 4 seconds. Determine:
a) the average net force on the car during this time
b) how far the car traveled while slowing down.
𝟓𝒎 𝟐𝟎𝒎
−
a) F = ma We need acceleration to get the force. 𝒂 = 𝒔 𝟒𝒔 𝒔 =-3.75m/s2
F = 2000kg(-3.75m/s2) = -7500N
b)
V2 =
V2 +
2aΔx  ∆𝒙 =
𝑽𝟐𝒇 −𝑽𝟐𝒊
𝟐𝒂
=
(𝟓𝒎/𝒔)𝟐 −(𝟐𝟎𝒎/𝒔)𝟐
𝟕𝟓𝒎
𝟐(−𝟑. 𝟐 )
𝒔
=50m
16. The kids are being pulled with a 120N force. The rope is 25 ̊ above the
horizontal, the mass of kids and sled is 100 kg, and the frictional force
is 15N. What is the acceleration of the sled?
We have all of the forces in the direction of acceleration (x-direction),
so the next step is to sum the forces in the direction of acceleration.
-Fk + Fpx = ma
−15𝑁+108.8𝑁
𝑎=
= 0.94m/s2
100𝑘𝑔
Notice how we didn’t need the normal force.
Fk = 15N
25o
Fpx = 120Ncos25 = 108.8N
Fg = 100kg(9.8m/s2) = 980N
17. A 60kg skier, starting from rest, is moving down a hill at a 35 ̊ angle. The
coefficient of friction is 0.08. What is the acceleration of the skier?
We need to find the normal force, so draw force diagram, sum the forces in
the y-direction, set equal to 0 (not accelerating in the vertical direction) and
solve for the normal force.
Fg = mg = 60kg(9.8m/s2) = 588N
–Fgx + FN = 0  FN = 588Ncos35 = 481.7
Now plug the normal force into the friction equation.
Fk = µFN = 0.08(481.7N) = 38.5N
Now sum the forces in the direction of acceleration (x-direction)
-Fk + Fgx = ma
35o
𝒂=
−𝟑𝟖.𝟓𝑵+𝟓𝟖𝟖𝑵𝒔𝒊𝒏𝟑𝟓
©Modeling Workshop Project 2006
𝟔𝟎𝒌𝒈
=4.98m/s2
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Unit V ws4 v3.0