Transcript Newton’s Second Law

Science Starter!
Draw a free-body diagram for:
1) A chair at rest on the floor.
2) A ball rolling to the right and slowing down
across a grassy field.
1.)
2.)
Remember that….
according to Newton’s 1st Law:
- objects at rest remain at rest unless a
force is applied to move them
- objects in motion stay in motion unless a
force is applied to change their speed or
direction
Newton’s Second Law
The net force applied to an object to change its state
of motion is directly proportional to the object’s mass
and resulting acceleration of the object.
Newton’s 2nd Law (formula form)
FNet = ma
Fnet (or ΣF) - “Net Force”: The vector sum of all forces
acting on an object [Newtons (N)]
m - mass (kg): constant value, depends on amount of
matter in the object
a: acceleration (m/s2)
Weight vs. Mass
Weight: “Fg” force of gravity pulling on an
object
F = ma
a  acceleration
“g”  acceleration due to gravity (9.8 m/s2)
THEREFORE:
Fg = mg
Example
A ball has a mass of 8.0 kg. What is the ball’s
weight on earth ?
(1)
m = 8.0 kg
g = 9.8 m/s2
Fg = ?
(2) Fg = mg
(3) Fg = (8.0 kg) (9.8 m/s2)
(4-5) Fg = 78.4 N
Net Force
Forces in “x” direction can be added,
Forces in “y” direction can be added:
CHEAT: Direction of acceleration is the
POSITIVE DIRECTION
Can be written as:
FNETx OR Σ Fx
FNETy OR Σ Fy
Example 1
A 10 kg wagon is being pulled to the right with
20 N of force. Between the tires and the
ground, there is 12 N of friction.
a) Draw a free-body diagram.
b) Write an “FNET Equation” for the vertical and
horizontal directions.
c) Determine the acceleration of the wagon.
Σ F y = FN – F g
(0) = FN – (10)(9.8)
Σ Fx = F – Ff
m(a) = F – Ff
10 (a) = (20) – (12)
10 (a) = 8
a = 0.8 m/s2
Example 2 – Think!
Two students are playing tug-of-war over a 15 kg
crate. Joe pulls to the right with 40 N of force,
while Bob pulls to the left with 60 N of force.
The frictional force between the crate and the
ground is 5 N.
a) Draw a free-body diagram.
b) Write an “FNET Equation” for the vertical
and horizontal directions.
c) Determine the acceleration of the crate.
Σ F y = FN – F g
Σ Fx = FBob – Ff – FJoe
(0) = FN – (15)(9.8) m(a) = FBob – Ff – FJoe
15 (a) = (60) – (5) – (40)
15 (a) = 15
a = 1.0 m/s2