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Physics 1
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NEWTON’S LAW OF MOTION
FORCE (F)
Described as either push or pull that can cause a mass (body) to accelerate (cause of motion)
A vector quantity that is the product of mass (m) & acceleration (a).
F = m a
In MKS it is in unit : Newtons (N) , 1 N = 1 kg-m/s2
In CGS it is in unit : dynes , 1 dyne = 1 g-cm/s2
In English Units : Pound (₤ or lbs) , Pound-force (lbf)
CONVERSION :
1 dyne = 1x10-5 N = 10-5 N
1 lbs = 4.448 N
COMPONENTS
m
F = 5 N
θ
m Fx = 5 N cos θ
θ
Fy = 5 N sin θ
=
FORCE (F)
1. Force due to Gravity / Weight FGRAV = W = mg, where g = 9.8 m/s2 or 32 ft/s2
m
W = mg W = mg θ
Weight is ALWAYS directed TOWARDS (attractive to) the earth, even if the surface is at an angle.
Weight is ALWAYS acting on the body
KINDS OF FORCE (F)
KINDS OF FORCES 2. Longitudinal Forces
Forces acting along the length of an object
Common in ropes, cables, solid cylinders
TENSION (T)
Pull Force on an object. Its end effect is to STRETCH an object.
COMPRESSION (C)
Push Force on an object. Its end effect is to FLATEN an object.
- Compression usually is due to normal forces between two objects in contact
C C
T T
Length (L)
Length (L)
KINDS OF FORCES 3. Contact Forces
Forces due to interaction between different surfaces
FRICTIONAL FORCE / FRICTION (f)
Force that oppose motion of an object.
Always parallel to the contact surface & directed opposite the motion of the object
NORMAL FORCE (Ŋ or N)
Reaction Force due to Weight of the object(s) in contact
Always Perpendicular to the contact surface
m
W = mg N
f
W = mg θ
f
KINDS OF FORCES 3. Contact Forces
f α N
f = μ N
μ – Coefficient of Friction
μS – Coefficient of Static Friction
μK – Coefficient of Kinetic Friction
μS = tan θf , (This θ MUST be the Angle of friction or repose)
KINDS OF FORCES 3. Contact Forces
Equations for Normal Force
m
W = mg W = mg
θ
N
N
W = mg
β
N
θ
F
N = W = mg N = Wy = W cosθ
N = mgcosθ
N = Wy – Fy
N = mgcosβ – Fsinθ
F
KINDS OF FORCES 3. Contact Forces
Equations for Frictional Force
m
W = mg W = mg
θ
N
N
W = mg
β
N
θ
F
N = W = mg N = Wy = W cosθ
N = mg cosθ
N = Wy – Fy
N = mgcosβ – F sinθ
F
f
f = μN = μmg
f
f
f = μN = μmg cosθ f = μN
f = μ(mgcosβ – F sinθ)
NEWTON’S FIRST LAW OF MOTION
“A body acted on by NO net force either stays motionless or moves, but with constant velocity
and zero acceleration”
NEWTON’S LAW OF MOTION
NEWTON’S SECOND LAW OF MOTION
NEWTON’S LAW OF MOTION
“A body requires a net force to accelerate” “The acceleration is directly proportional to the net force but
inversely proportional to the body’s mass”
“The direction of the net force is the same as the direction of the acceleration”
NEWTON’S THIRD LAW OF MOTION
“To every action there is always opposed an equal reaction, same in magnitude but opposite
in direction.”
m
W = mg
N
m
W = mg
T
NEWTON’S LAW OF MOTION
EQUILIBRIUM – The effects of all forces acting on a single point on the body cancel one another. There is no change in motion
An object or body is at Equilibrium, when :
(1)The body stays at rest
OR
(2)The body moves in a straight line but in constant or uniform velocity (No acceleration)
For Both :
ΣFx = 0 & ΣFy = 0
Therefore : R = 0 or Fnet = 0
FIRST CONDITION OF EQUILIBRIUM (FCE)
FREE BODY DIAGRAM (FBD)
Used to analyze forces acting on a body by isolating the body with all forces acting on it.
Steps
1. A. Represent the object as a point mass
B. If it is a system (multiple objects), a certain point on it will be indicated for reference, use this as the “point mass”.
2. Draw ALL forces ACTING ON the body from this point
Internal – Weight External – Applied Force, Friction Normal Force (Coming from Other Body in contact), Tension (Always away from the body)
3. Don’t forget to show also the angle of the force (if angled).
WRONG FBD CORRECT FBD
W = mg W = mg
f f
N
N
θ
F
m
W = mg N
f
θ
F
θ
F
TIPS ON FBD
DON’T Draw Vectors GOING towards the point mass (even though it may show in the figure) draw it away from the point mass, it does the same thing.
EXAMPLE 1
EXAMPLE 2
W = mg θ
f
WRONG FBD * CORRECT FBD *
W = mg W = mg
f f
N θ θ
N
TIPS ON FBD
PULLEYS & Weights Pulleys are analyzed as frictionless & of negligible weight.
For FCE : Tension of rope or cable passing through the pulley is equal to the weights HANGING from them.
T = W
θ
2 kg
W = mg
T = W
T = W T = W
2 kg
W1 = mg
T = W1
T = W1
T = W1
T = W1
TIPS ON FBD
Using the inclined surface as the x-axis (rotate of axis)
W = mg θ
f
1000 lbs
45°
60°
O
Must have at least 1 pair of perpendicular forces and 90° angle is visible.
ALLOWED since f & N are Perpendicular
NOT ALLOWED :No perpendicular forces or angles with respect to point O, thus not practical to use any inclined as x-axis.
TIPS ON FBD
TIPS ON FBD
If a system consists of two or more objects. Multiple FBD’s may be required :
A
B P = ? A
FIRST CONDITION OF EQUILIBRIUM (FCE)
Sample Problems :
1. A 5 kg block will start to slide down at constant speed from a surface when it is inclined at 40° with the horizontal. Determine the Frictional force, Normal force and the coefficient of static friction. 40°
Solution : Draw the forces acting on the body
f N
Draw the FBD
W = mg
f N
40°
40° W = mg
Since we have perpendicular forces along the inclined (f & N) we can use the inclined as our x-axis
Re-draw the FBD
W = mg
f
N
40°
Draw the component vectors of angled vectors
W
f
N
40°
Wx
Wy
FIRST CONDITION OF EQUILIBRIUM (FCE)
W
f
N
40°
Wx
Wy
ΣFx = 0 → (+)
Fnet = 0
− Wx + f = 0
f = Wx = Wsinθ
ΣFy = 0↑ (+)
+ N – Wy = 0
N = Wy = Wcosθ
N = mg cos θ
N = (5kg)(9.8m/s2)(cos 40°)
N = 37.54 N
Friction and Normal force and μS
FIRST CONDITION OF EQUILIBRIUM (FCE)
f = mg sinθ
f = (5kg)(9.8m/s2)(sin 40°)
f = 31.5 N
f = μ N
μ = f/N μS = f/N = 31.5 N / 37.54 N μS = 0.839
FIRST CONDITION OF EQUILIBRIUM (FCE)
2. A woman at an airport is pulling a 15 kg suitcase (with wheels) at constant speed of 2 m/s by pulling on the handle attached to the bag (this makes an angle θ above the horizontal). She pulls with a 56 N force, and the frictional force is 20N. What is the angle “θ”, the normal force & the coefficient of kinetic friction?
F = 56 N
m = 15 kg
f = 20 N
θ N
W = mg
F = 56 N
θ f = 20 N
N
W = mg
Fy
θ f = 20 N
Fx
ΣFx = 0 → (+)
Fnet = 0
− f + Fx = 0
f = Fx
ΣFy = 0↑ (+)
+ N + Fy – W = 0
N = W – Fy
N = mg – F sin θ
N = (15kg)(9.8m/s2) – (56N)(sin 69.08°)
N = 94.69 N
Solving for θ
FIRST CONDITION OF EQUILIBRIUM (FCE)
f = F cosθ
20 N = 56 N(cos θ)
θ = 69.08°
f = μ N
μ = f/N μK = f/N = 20 N / 94.69 N μK = 0.211
N
W = mg
Fy
θ f = 20 N
Fx
,Normal force and μK
FIRST CONDITION OF EQUILIBRIUM (FCE)
3. A wet shirt weighs 4 N. It is hanged to dry on a metal clothesline. The shirt is placed at the very center of the length of the clothesline, and the angle formed with respect to the horizontal due to the weight of the shirt on either side are equal. What are the tensions on each side of the clothesline?
θ θ
W = 4 N
T2
θ θ
T1
W = 4 N
T2y
θ θ
T1y
T1x T2x
2 m
0.4 m
θ
1 m
0.4 m tan θ = (0.4 m)/(1 m)
θ = 21.8°
ΣFx = 0 → (+)
Fnet = 0
− T1x + T2x = 0
T2x = T1x
ΣFy = 0↑ (+)
+ T1y + T2y – W = 0
T1 sinθ + T2 sinθ = W
T1 sinθ + T1 sinθ = W
Solving for Tensions
FIRST CONDITION OF EQUILIBRIUM (FCE)
T2 cosθ = T1cosθ
T2 = T1 W = 4 N
T2y
θ θ
T1y
T1x T2x
2(T1 sinθ) = W
T1 = W/ (2sinθ)
T1 = [(4N)/[2sin(21.8°)]
T1 =5.385N T2 = 5.385N
NEWTON’S SECOND LAW OF MOTION
NEWTON’S LAW OF MOTION
“A body requires a net force to accelerate” “The acceleration is directly proportional to the net force but
inversely proportional to the body’s mass”
“The direction of the net force is the same as the direction of the acceleration”
m
Body of mass “m” at rest on a frictionless surface
m
Due to net force “Fnet” going to the left the object will accelerate also to the left
Fnet
a Fnet = ma
m F
a
W = mg N
a) ΣFx = max & ΣFy = may
b) ΣFx = max & ΣFy = 0
c) ΣFx = 0 & ΣFy = may
If Fnet = ma
Possibilities :
Purely Horizontal Movement
Purely Vertical Movement
NEWTON’S SECOND LAW OF MOTION (NSLM)
NET FORCES
Net Force due to Contact Forces
m
W = mg W = mg
θ
N
N
W = mg
β
N
θ
F
N = W = mg N = Wy = W cosθ
N = mg cosθ
N = Wy – Fy
N = mgcosβ – F sinθ
F
f
Fnet = F – f
f
Fnet = Wx – f
Fnet = Wsinθ – f
Fnet = mg sinθ – f
f
Fnet = Fx – Wx – f
Fnet = F cos θ – W sinβ – f
Fnet = F cosθ - mg sinβ – f
NEWTON’S SECOND LAW OF MOTION (NSLM)
Sample Problems :
1. A 5 kg block slides down a plane inclined at 40° to the horizontal. Find the acceleration of the block
a) If the plane is frictionless b) If the coefficient of kinetic friction is
0.20 40°
Solution : Draw the forces acting on the body
f N
Draw the FBD
W = mg
f N
40°
40° W = mg
NEWTON’S SECOND LAW OF MOTION (NSLM)
Since we have perpendicular forces along the inclined (f & N) we can use the inclined as our x-axis
Re-draw the FBD
W = mg
f
N
40°
a
Draw the component vectors of angled vectors
W
f
N
40°
a
Wx
Wy
ΣFx = max → (+)
Fnet = ma
− Wx + f = − ma
ma = Wx – f
ΣFy = 0↑ (+)
+ N – Wy = 0
N = Wy = Wcosθ
N = mg cos θ ma = mg sinθ – f
f = μ N
W
f
N
40°
Wx
Wy
a
ma = mg sinθ - μN
ma = mg sinθ – μmg cosθ ma = mg (sin θ – μ cosθ)
NEWTON’S SECOND LAW OF MOTION (NSLM)
a = g (sin θ – μ cosθ)
1. A 5 kg block slides down a plane inclined at 40° to the horizontal. Find the acceleration of the block
a) If the plane is frictionless b) If the coefficient of kinetic friction is 0.20
NEWTON’S SECOND LAW OF MOTION (NSLM)
(a) a = ? If f = 0, hence μ = 0
a = g [ (sin 40°) − μ ( cos 40°) ]
a = (9.8 m/s2) [ (sin 40°) − 0 ( cos 40°) ]
a = 6.3 m/s2
(b) a = ? If μ = 0.2 a = g [ (sin 40°) − μ ( cos 40°) ]
a = (9.8 m/s2) [ (sin 40°) − 0.2 ( cos 40°) ]
a = 4.8 m/s2
2. A car (2,000 kg) is traveling at 28.7 m/s when the driver locks the breaks to stop the car. What will be the shortest distance ( from the point where the breaks were locked up to the full stopping point), if the coefficient of kinetic friction between the tires and pavement (road) is 0.8?
NEWTON’S SECOND LAW OF MOTION (NSLM)
VO = 28.7 m/s
s
VF = 0 a
Using Kinematics Eq’n (3)
VF2 = VO
2 + 2as
s = (VF2 − VO
2)/(2a)
NEWTON’S SECOND LAW OF MOTION (NSLM)
f
W = mg
N
2. A car (2,000 kg) is traveling at 28.7 m/s when the driver locks the breaks to stop the car. What will be the shortest distance ( from the point where the breaks were locked up to the full stopping point), if the coefficient of kinetic friction between the tires and pavement (road) is 0.8?
a f
W = mg
N
a
Using NSLM to determine the acceleration FBD :
NEWTON’S SECOND LAW OF MOTION (NSLM)
2. A car (2,000 kg) is traveling at 28.7 m/s when the driver locks the breaks to stop the car. What will be the shortest distance ( from the point where the breaks were locked up to the full stopping point), if the coefficient of kinetic friction between the tires and pavement (road) is 0.8?
f
W = mg
N
a ΣFx = max → (+)
Fnet = ma
− f = + ma ΣFy = 0↑ (+)
+ N – W = 0
N = W
N = mg f = μ N
– μN = +ma – μmg = +ma – μg = +a
a = – μg = − (0.8)(9.8 m/s2) = − 7.84 m/s2
a = 7.84 m/s2, deceleration
2. A car (2,000 kg) is traveling at 28.7 m/s when the driver locks the breaks to stop the car. What will be the shortest distance ( from the point where the breaks were locked up to the full stopping point), if the coefficient of kinetic friction between the tires and pavement (road) is 0.8?
NEWTON’S SECOND LAW OF MOTION (NSLM)
VO = 28.7 m/s
s
VF = 0 a
Using Kinematics Eq’n (3)
VF2 = VO
2 + 2as
s = (VF2 − VO
2)/(2a)
s = [(02 − (28.7m/s)2]/[(2)(−7.84 m/s2)]
s = 52.53 m
a = 7.84 m/s2, deceleration
s = (−823.69 m2/s2)/(−15.68 m/s2)
NEWTON’S SECOND LAW OF MOTION (NSLM)
3. A 3.5 kg pail is dropped into a 15 m empty deep well, starting from rest at the top. The tension in the rope is constant at 14.8 N as the pail drops. What is the time to reach the bottom of the well?
h = 15 m
VO = 0 Using Kinematics Eq’n (2)
h = VOt + ½ at2
Where a ≠ g
NEWTON’S SECOND LAW OF MOTION (NSLM)
3. A 3.5 kg pail is dropped into a 15 m empty deep well, starting from rest at the top. The tension in the rope is constant at 14.8 N as the pail drops. What is the time to reach the bottom of the well?
Using NSLM to determine the acceleration FBD :
T = 14.8 N
W = mg
a
W = mg
T = 14.8 N
a
NEWTON’S SECOND LAW OF MOTION (NSLM)
3. A 3.5 kg pail is dropped into a 15 m empty deep well, starting from rest at the top. The tension in the rope is constant at 14.8 N as the pail drops. What is the time to reach the bottom of the well?
W = mg
T = 14.8 N
a
Fnet = ma
ΣFx = 0 → (+) ΣFy = may↑ (+)
+ T – W = – ma
ma = mg – T
a = ( mg – T ) / m
a = [(3.5kg)(9.8 m/s2) – 14.8 N]/(3.5kg)
a = 5.57 m/s2 , (downward)
NEWTON’S SECOND LAW OF MOTION (NSLM)
3. A 3.5 kg pail is dropped into a 15 m empty deep well, starting from rest at the top. The tension in the rope is constant at 14.8 N as the pail drops. What is the time to reach the bottom of the well?
h = 15 m
VO = 0 Using Kinematics Eq’n (2)
h = VOt + ½ at2
Where a ≠ g
a = 5.57 m/s2 , (downward)
– 15m = (0)t + ½(– 5.57 m/s2)t2
– 15m = (– 2.785 m/s2)t2
t2 = (15m/ 2.785 m/s2)
t2 = 5.386s2
t = 2.32 s
Prob 4 : Given :
System is released from rest. Determine the acceleration of the system, when it is already in motion
m1 = 2 kg
m2
μS = 0.65 μK = 0.2
a = ?
FBD of m1
W1 = m1g
N1
T f
FBD of m2
W2 = m2g
T
a
a
Derive first the equation for the acceleration
Use NSLM on the first FBD (m1)
ΣFx = max → (+)
Fnet = ma
− f + T = + m1a
ΣFy = 0↑ (+)
+ N1 – W1 = 0
N1 = W1
N1 = m1g f = μ N
– μN1 + T = m1a – μm1g + T = m1a We solve for “T” because it is the common force between m1 & m2
T = m1a + μm1g (eq’n 1)
System is released from rest. Determine the acceleration of the system, when it is already in motion
m1 = 2 kg
m2
μS = 0.65 μK = 0.2
a = ?
FBD of m1
W1 = m1g
N1
T f
FBD of m2
W2 = m2g
T
a
a
Use NSLM on the second FBD (m2)
ΣFx = 0 → (+)
Fnet = ma
ΣFy = may↑ (+)
T – W2 = – m2a
T = W2 – m2a
T = m2g – m2a (eq’n 2)
Prob 4 : Given :
System is released from rest. Determine the acceleration of the system, when it is already in motion
m1 = 2 kg
m2
μS = 0.65 μK = 0.2
a = ?
FBD of m1
W1 = m1g
N1
T f
FBD of m2
W2 = m2g
T
a
a
m2g – m2a
(eq’n 1 = eq’n 2)
m1a + μm1g =
m1a + m2a = m2g – μm1g
a (m1 + m2) = g (m2 – μm1)
a = g (m2 – μm1)
(m1 + m2)
Prob 4 : Given :
System is released from rest. Determine the acceleration of the system, when it is already in motion
m1 = 2 kg
m2
μS = 0.65 μK = 0.2
a = ?
FBD of m1
W1 = m1g
N1
T f
FBD of m2
W2 = m2g
T
a
a
a = g (m2 – μm1)
(m1 + m2)
Solve first for the value of m2
When the object is about to move the frictional coefficient is static and the acceleration is zero
0 = g (m2 – μSm1)
(m1 + m2)
0 = m2 – μSm1 The equation is reduced to :
m2 = μSm1 = (0.65)(2kg)
m2 = 1.3 kg
Prob 4 : Given :
System is released from rest. Determine the acceleration of the system, when it is already in motion
m1 = 2 kg
m2
μS = 0.65 μK = 0.2
a = ?
FBD of m1
W1 = m1g
N1
T f
FBD of m2
W2 = m2g
T
a
a
a = g (m2 – μKm1)
(m1 + m2)
Next we solve for acceleration using µK and m2
a = 9.8m/s2 [1.3kg – (0.20)(2kg)]
(2kg + 1.3kg)
a = 2.673 m/s2
Prob 4 : Given :
A
B
Consider the figure shown below. Block A weighs 50N and block B weighs 27N . Once block B is set into downward motion, it descends at a constant speed. a)Calculate the coefficient of kinetic friction between block A and the table top. b)A cat, of weight 50N , jumps on top of block B. If block B is now set in to downward motion, what is its acceleration?
WA = 50N
WB = 27N
This has the very same procedure in derivation for acceleration using NSLM as problem 1
a = g (mB – μmA)
(mA + mB)
The working equation is still
Prob 5 : Given :
A
B
Consider the figure shown below. Block A weighs 50N and block B weighs 27N . Once block B is set into downward motion, it descends at a constant speed. a)Calculate the coefficient of kinetic friction between block A and the table top. b)A cat, of weight 50N , jumps on top of block B. If block B is now set in to downward motion, what is its acceleration?
WA = 50N
WB = 27N
a) Solve for µK
a = g (mB – μKmA)
(mA + mB)
Condition : Block B (& the system) moves at constant speed (a = 0) downward
Prob 5 : Given :
0 = g (mB – μKmA)
(mA + mB)
0 = mB – μKmA
µK = mB/mA
µK = (WB/g)/(WA/g) = (WB/(WA)
µK = (27N/50N)
µK = 0.54
A
B
Consider the figure shown below. Block A weighs 50N and block B weighs 27N . Once block B is set into downward motion, it descends at a constant speed. a)Calculate the coefficient of kinetic friction between block A and the table top. b)A cat, of weight 50N , jumps on top of block B. If block B is now set in to downward motion, what is its acceleration?
WA = 50N
WB = 27N+50N
b) Solve for “a”
a = g (mB – μKmA)
(mA + mB)
Condition : Block B now has additional weight due to the cat, the system will now accelerate
Prob 5 : Given :
mA = WA/g = (50N/9.8m/s2)
mB = WB/g = (27N+50N)/(9.8m/s2)
a = 9.8m/s2 [7.857kg – (0.54)(5.102kg)]
(5.102kg + 7.857kg)
mB = 7.857 kg
a = 3.86 m/s2
mA = 5.102 kg
a
Consider the figure shown below. Block A is 1kg and block B is 2kg. The inclined of block A is 50° with the horizontal while that of block B is 25°. In what direction will the system go and what is the acceleration? Assume that the surface is frictionless.
mA = 1 kg
mB = 2 kg
θA = 50° θB = 25° FBD of mA
WA = mAg
NA T a
θA = 50°
WA = mAg
NA
T
a
θA = 50°
WAY = WAcosθA
NA
T
a
WAX = WAsinθA
a
θA = 50°
Assume direction is to the left. (This is just an assumed or guess direction)
Prob 6: Given :
Consider the figure shown below. Block A is 1kg and block B is 2kg. The inclined of block A is 50° with the horizontal while that of block B is 25°. In what direction will the system go and what is the acceleration? Assume that the surface is frictionless.
mA = 1 kg
mB = 2 kg
FBD of mA
WAY = WAcosθ
NA
T
a
WAX = WAsinθ
a
ΣFx = max → (+)
Fnet = ma
ΣFy = 0↑ (+)
– WAX + T = – mAa
– WAsinθA + T = – mAa
T = WAsinθA – mAa (eq’n 1)
θA = 50° θB = 25°
θA = 50°
Prob 6 : Given :
Consider the figure shown below. Block A is 1kg and block B is 2kg. The inclined of block A is 50° with the horizontal while that of block B is 25°. In what direction will the system go and what is the acceleration? Assume that the surface is frictionless.
mA = 1 kg
mB = 2 kg
FBD of mB
NB
T
a
θB = 25°
WB = mBg
NB
T
a
θB = 25° WBY = WBcosθB
NB
T
a
WBX = WBsinθB
a
θA = 50° θB = 25°
θB = 25°
Prob 6: Given :
Consider the figure shown below. Block A is 1kg and block B is 2kg. The inclined of block A is 50° with the horizontal while that of block B is 25°. In what direction will the system go and what is the acceleration? Assume that the surface is frictionless.
mA = 1 kg
mB = 2 kg
FBD of mB
WBY = WBcosθB
NB
T
a
WBX = WBsinθB
a
θA = 50° θB = 25°
θB = 25°
ΣFx = max → (+)
Fnet = ma
ΣFy = 0↑ (+)
– T + WBX = – mBa
− T + WBsinθB = – mBa
T = WBsinθB + mBa (eq’n 2)
Prob 6 : Given :
Consider the figure shown below. Block A is 1kg and block B is 2kg. The inclined of block A is 50° with the horizontal while that of block B is 25°. In what direction will the system go and what is the acceleration? Assume that the surface is frictionless.
mA = 1 kg
mB = 2 kg
a
θA = 50° θB = 25°
T = WBsinθB + mBa (eq’n 2) T = WAsinθA – mAa (eq’n 1)
(mAg)sinθA – mAa = (mBg)sinθB + mBa
(eq’n 1) = (eq’n2)
(mAg)sinθA – (mBg)sinθB = mAa + mBa
g(mAsinθA – mBsinθB) = a(mA+ mB)
g(mAsinθA – mBsinθB)
(mA+ mB) a =
Substitute all given values :
a = – 0.259m/s2 The negative sign means that the assumed direction of “a” was wrong. Hence it should be to the right :
a = 0.259m/s2 to the right
Prob 6 : Given :