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NEWTON’S LESSON 6
FINDING ACCELERATION
In Lesson 2, we learned how to determine the net force if the magnitudes of all the
individual forces are known. In this lesson, we will learn how to determine the
acceleration of an object if the magnitudes of all the individual forces are known. The
three major equations which will be useful are the equation for net force (Fnet = m*a),
the equation for gravitational force (Fgrav = m*g), and the equation for frictional force
(Ffrict = "mu"*Fnorm).
The process of determining the acceleration of an object demands that the mass and the
net force are known.
The net force is the vector sum of all the individual forces.
The three major equations which will be useful are the equation for net force (Fnet =
m*a), the equation for gravitational force (Fgrav = m*g), and the equation for frictional
force (Ffrict = μ*Fnorm).
If mass (m) and net force (Fnet) are known, then the acceleration is determined by use of
the equation.
Thus, the task involves using the above equations, the given information, and your
understanding of Newton's laws to determine the acceleration.
EXAMPLE: An applied force of 50 N is used to accelerate an object to the right across a
frictional surface. The object encounters 10 N of friction. Use the diagram to determine
the normal force, the net force, the mass, and the acceleration of the object. (Neglect air
resistance.)
EXAMPLE: An applied force of 20 N is used to accelerate an object to the
right across a frictional surface. The object encounters 10 N of friction. Use the
diagram to determine the normal force, the net force, the coefficient of friction
("mu") between the object and the surface, the mass, and the acceleration of the
object. (Neglect air resistance.)
EXAMPLE: A 5-kg object is sliding to the right and encountering a friction force which
slows it down. The coefficient of friction ("μ") between the object and the surface is 0.1.
Determine the force of gravity, the normal force, the force of friction, the net force, and
the acceleration. (Neglect air resistance.)
FREE FALL AND TERMINAL VELOCITY
Falling with Air Resistance
As an object falls through air, it usually encounters some degree of air resistance.
Air resistance is the result of collisions of the object's leading surface with air molecules.
The actual amount of air resistance encountered by the object is dependent upon a variety
of factors:
the speed of the object and
the cross-sectional area of the object.
Increased speeds result in an increased amount of air resistance. Increased cross-sectional
areas result in an increased amount of air resistance.
ANIMATION: ELEPHANT AND FEATHER AIR RESISTANCE
What falls faster, an elephant or a feather? Why?
Why does an object which encounters air resistance eventually reach a terminal velocity?
To answer this questions, Newton's second law will be applied to the motion of a falling
85-kg skydiver. Determine the net force and acceleration of the skydiver at each
instance.
As an object falls, it picks up speed. The increase in speed leads to an increase in the
amount of air resistance.
Eventually, the force of air resistance becomes large enough to balances the force of
gravity. At this instant in time, the net force is 0 Newtons; the object will stop
accelerating. The object is said to have reached a terminal velocity.
ANIMATION- Skydiving
In situations in which there is air resistance, more massive objects fall faster than less
massive objects. But why?
To answer the why question, it is necessary to consider the free-body diagrams for objects
of different mass. Consider the falling motion of two skydivers: one with a mass of 100
kg (skydiver plus parachute) and the other with a mass of 150 kg (skydiver plus
parachute). The free-body diagrams are shown below for the instant in time in which they
have reached terminal velocity.
As learned above, the amount of air resistance depends upon the speed of the object. A
falling object will continue to accelerate to higher speeds until they encounter an amount
of air resistance which is equal to their weight. Since the 150-kg skydiver weighs more
(experiences a greater force of gravity), it will accelerate to higher speeds before reaching
a terminal velocity.
Thus, more massive objects fall faster than less massive objects because they are acted
upon by a larger force of gravity; for this reason, they accelerate to higher speeds until
the air resistance force equals the gravity force.
EXAMPLE:
A 72-kg skydiver is falling from 10000 feet. At an instant during the fall,
the skydiver encounters an air resistance force of 540 Newtons. Determine
the acceleration of the skydiver at this instant.
There are two forces acting upon the skydiver - gravity (down) and air resistance
(up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. The
sum of the vertical forces is
·Fy = 540 N, up + 706 N, down = 166 N, down
The acceleration of the skydiver can be computed using the equation ·Fy = m•ay.
ay = (166 N, down) / (72 kg) = 2.30 m/s/s, down
NEWTON’S LESSON 6 HOMEWORK
1. Which one(s) of the following force diagrams depict an object moving to the right
with a constant speed? List all that apply.
2. Edwardo applies a 4.25-N rightward force to a 0.765-kg book to accelerate it across a
table top. The coefficient of friction between the book and the tabletop is 0.410.
Determine the acceleration of the book.
3. In a physics lab, Kate and Rob use a hanging mass and pulley system to exert a 2.45 N
rightward force on a 0.500-kg cart to accelerate it across a low-friction track. If the total
resistance force to the motion of the cart is 0.720 N, then what is the cart's acceleration?
4. A 72.0-kg skydiver is falling from 10 000.0 feet. After reaching terminal velocity, the
skydiver opens his parachute. Shortly thereafter, there is an an instant in time in which
the skydiver encounters an air resistance force of 1180 Newtons. Determine the
acceleration of the skydiver at this instant.
5. Free fall is motion in which gravity is the only force acting. (a) Is a sky diver who
has reached terminal speed in free fall?
6. A 4.44-kg bucket suspended by a rope is accelerated upwards from an initial rest
position. If the tension in the rope is a constant value of 83.1 Newtons, then determine the
acceleration (in m/s/s) of the bucket.
7. A shopper in a supermarket pushes a loaded cart with a horizontal force of 16.5
Newtons. If the cart has a mass of 33.8 kg, how far (in meters) will it move in 1.31
seconds, starting from rest? (Neglect resistive forces.)
8. A tow truck exerts a 18300-N force upon a 1200.0-kg car to drag it out of a mud
puddle onto the shoulder of a road. A 17900 N force opposes the car's motion. The plane
of motion of the car is horizontal. Determine the time required to drag the car a distance
of 6.90 meters from its rest position.
HOMEWORK KEY
1.
2.
3.
4.
5.
6.
7.
8.
A,C
1.54 m/s2, right
3.46 m/ s2, right
6.59 m/s2, up
no key
8.92 m/s2, up
0.419 m
6.43 s
NEWTON’S LESSON 6 HOMEWORK
1. Which one(s) of the following force diagrams depict an object moving to the right
with a constant speed? List all that apply.
Answer: AC
If an object is moving at a constant speed in a constant rightward direction, then the
acceleration is zero and the net force must be zero. Choice B and E show a rightward net
force and therefore a rightward acceleration, inconsistent with the described motion.
2. Edwardo applies a 4.25-N rightward force to a 0.765-kg book to accelerate it across a
table top. The coefficient of friction between the book and the tabletop is 0.410.
Determine the acceleration of the book.
Answer: Fgrav = 7.50 N; Fnorm = 7.50 N; Ffrict = 3.07 N; Fnet = 1.18 N, right; a = 1.54
m/s/s, right
The starting point for any problem such as this is the construction of a
free-body diagram in which you show all the individual forces which are
acting upon the book. There are two vertical forces - gravity and normal
force. There are two horizontal forces - friction and the applied force.
Since there is no vertical acceleration, normal force = gravity force. Each
of these forces can be determined using the equation Fgrav = m • g = (0.765
kg) • (9.8 m/s/s) = 7.497 N
The force of friction can be determined using the equation Ffrict = mu •
Fnorm. So Ffrict = (0.410) • (7.497 N) = (3.0737 ... N)
The Fnet is the vector sum of all the forces: 4.25 N, right plus 3.0737 ... N,
left = 1.176... N, right.
Finally, a = Fnet / m = (1.176... N) / (0.765 kg) = 1.54 m/s/s.
3. In a physics lab, Kate and Rob use a hanging mass and pulley system to exert a 2.45 N
rightward force on a 0.500-kg cart to accelerate it across a low-friction track. If the total
resistance force to the motion of the cart is 0.720 N, then what is the cart's acceleration?
Answer: Fgrav = 4.90 N; Fnorm = 4.90 N; Fnet = 1.73 N, right; a = 3.46 m/s/s, right
The starting point for any problem such as this is the construction of a
free-body diagram in which you show all the individual forces which are
acting upon the book. There are two vertical forces - gravity and normal
force. There are two horizontal forces - friction and the applied force.
Since there is no vertical acceleration, normal force = gravity force. Each
of these forces could be determined using the equation Fgrav = m • g =
(0.500 kg)•(9.8 m/s/s) = 4.90 N.
The Fnet is the vector sum of all the forces: 2.45 N, right plus 0.72 N, left =
1.73 N, right.
Finally, a = Fnet / m = (1.73. N) / (0.500 kg) = 3.46 m/s/s.
4. A 72.0-kg skydiver is falling from 10 000.0 feet. After reaching terminal velocity, the
skydiver opens his parachute. Shortly thereafter, there is an an instant in time in which
the skydiver encounters an air resistance force of 1180 Newtons. Determine the
acceleration of the skydiver at this instant.
Answer: Answer: 6.59 m/s/s, up
There are two forces acting upon the skydiver - gravity (down) and air
resistance (up). The force of gravity has a magnitude of m•g = (72 kg)
•(9.8 m/s/s) = 706 N. The sum of the vertical forces is
·Fy = 1180 N, up + 706 N, down = 474 N, up
The acceleration of the skydiver can be computed using the equation ·Fy
= m•ay.
ay = (474 N, up) / (72 kg) = 6.59 m/s/s, up
5. Free fall is motion in which gravity is the only force acting. (a) Is a sky diver who
has reached terminal speed in free fall?
Answer: No, because when the force of gravity equals the force of air resistance, the sky
diver will stop accelerating and will reach a terminal speed.
6. A 4.44-kg bucket suspended by a rope is accelerated upwards from an initial rest
position. If the tension in the rope is a constant value of 83.1 Newtons, then determine the
acceleration (in m/s/s) of the bucket.
Answer: 14.2 m/s
There are two forces acting upon the bucket - the force of gravity (up) and the tension
force (down). The magnitude of the force of gravity is found from m•g; its
value is 43.5 N. These two forces can be summed as vectors to determine
the net force.
Fnet = ·Fy = 83.1 N, up + 43.5 N, down = 39.6 N, up
The acceleration can be calculated using Newton's second law of motion.
a = / m = (39.6 N, up) / (4.44 kg) = 8.92 m/s/s, up
7. A shopper in a supermarket pushes a loaded cart with a horizontal force of 16.5
Newtons. If the cart has a mass of 33.8 kg, how far (in meters) will it move in 1.31
seconds, starting from rest? (Neglect resistive forces.)
Answer: 0.419 m
Upon neglecting air resistance, there are three forces acting upon the object. The up and
down force balance each other and the acceleration is caused by the applied force. The
net force is 22.6 N, right (equal to the only rightward force - the applied force).
So the acceleration of the object can be computed using Newton's second law.
a = Fnet / m = (16.5 N, right) / (33.8 kg) = 0.488 m/s/s, right
The acceleration value can be used with other kinematic information (vi = 0 m/s,
t = 1.31 s) to calculate the final speed of the cart. The kinematic equation, substitution
and algebra steps are shown.
d = vi • t + 0.5 •a • t2
d = vi • t + + 0.5 • (0.488 m/s/s)•(1.31 s) 2
d = 0.419 m
8. A tow truck exerts a 18300-N force upon a 1200.0-kg car to drag it out of a mud
puddle onto the shoulder of a road. A 17900 N force opposes the car's motion. The plane
of motion of the car is horizontal. Determine the time required to drag
the car a distance of 6.90 meters from its rest position.
Answer: 6.43 s
Upon neglecting air resistance, there are four forces acting upon the object. The up and
down forces balance each other. The acceleration is rightward (or in the direction of the
applied force) since the rightward applied force is greater than the leftward friction force.
The horizontal forces can be summed as vectors in order to determine the net force.
Fnet = ·Fx = 18300 N, right - 17900 N, left = 400 N, right
The acceleration of the object can be computed using Newton's second law.
ax = ·Fx / m = (400 N, down) / (1200 kg) = 0.333 m/s/s, right
This acceleration value can be combined with other known kinematic information (vi = 0
m/s, d = 6.9 m) to determine the time required to drag the car a distance of 6.9 m. The
following kinematic equation is used; substitution and algebra steps are shown.
d = vi • t + 0.5 •a • t2
d = vi • t + 0.5 •a • t2
6.9 m = 0.5 • (0.333 m/s/s) • t2
6.9 m / (0.5 • 0.333 m/s/s ) = t2
41.4 = t2
6.43 s = t