Download Second

Weight: More Practice
Enter very large or very small numbers into your calculator using the EXP button:
5.356 EXP -26
Weight: More Practice
Weight: More Practice
Newton’s 2nd Law of Motion: Student
Learning Goal
The student will be able to state Newton’s 2nd Law
and apply it in qualitative and quantitative terms
to explain the effect of forces acting on objects.
(B3.4)
nd
2
Newton’s
Law of
Motion and Acceleration
Newton’s Second Law of Motion
If the net force is not zero, the object will be
accelerated in the direction of the net force.

 Fnet
a
m


Fnet  ma
Newton’s Second Law Example
Example: A weightlifter lifts a 165-kg weight by
exerting a force of 1.8 kN [up]. The force of
gravity on the object is 1.6 kN [down]. Draw a
FBD of the weight. What is the net force on the
weight? What is the acceleration of the weight?
Newton’s Second Law Example
1.8 kN
1.6 kN
Newton’s Second Law Example
1.8 kN
1.6 kN
The net force is 0.2 kN [up].
Newton’s Second Law Example
What is the acceleration of the weight?
Newton’s Second Law Example
What is the acceleration of the weight?
Givens : m  165 kg
Fnet  0.2 kN  0.2 1000 N  200 N
Unknown : a  ?
Newton’s Second Law Example
What is the acceleration of the weight?
Givens : m  165 kg
Fnet  0.2 kN  0.2 1000 N  200 N
Unknown : a  ?
Newton’s Second Law Example
What is the acceleration of the weight?
Givens : m  165 kg
Fnet  0.2 kN  0.2 1000 N  200 N
Unknown : a  ?
Fnet
Select : a 
m
200 N
Solve : a 
 1.2 kgN or1.2 sm2
165 kg
Newton’s Second Law Example
What is the acceleration of the weight?
Givens : m  165 kg
Fnet  0.2 kN  0.2 1000 N  200 N
Unknown : a  ?
Fnet
Select : a 
m
200 N
Solve : a 
 1.2 kgN or1.2 sm2
165 kg
Example with the Equations of
Motion
A car is travelling at 25 m/s [forward] when the
driver slams on the brakes and stops the car in
3.0 s. Calculate (a) the acceleration of the car
and (b) the net force needed to cause that
acceleration if the mass of the car is 1200 kg.
Example with the Equations of
Motion
A car is travelling at 25 m/s [forwards] when the
driver slams on the brakes and stops the car in
3.0 s. Calculate (a) the acceleration of the car.
Example with the Equations of
Motion
A car is travelling at 25 m/s [forwards] when the
driver slams on the brakes and stops the car in
3.0 s. Calculate (a) the acceleration of the car.
Givens : v1  25 ms
v2  0 ms
 t  3 .0 s
Unknown : a  ?
Example with the Equations of
Motion
A car is travelling at 25 m/s [forwards] when the
driver slams on the brakes and stops the car in
3.0 s. Calculate (a) the acceleration of the car.
Givens : v1  25 ms
v2  0 ms
 t  3 .0 s
Unknown : a  ?
v2  v1
Select : a 
t
0 ms  25 ms
Solve : a 
 8.333 sm2
3.0 s
Example with the Equations of
Motion
A car is travelling at 25 m/s [forwards] when the
driver slams on the brakes and stops the car in
3.0 s. Calculate (a) the acceleration of the car.
Givens : v1  25 ms
v2  0 ms
 t  3 .0 s
Unknown : a  ?
v2  v1
Select : a 
t
0 ms  25 ms
Solve : a 
 8.333 sm2
3.0 s
Example with the Equations of
Motion
A car is travelling at 25 m/s [forwards] when the
driver slams on the brakes and stops the car in
3.0 s. Calculate (a) the acceleration of the car.
Givens : v1  25 ms
v2  0 ms
 t  3 .0 s
Unknown : a  ?
v2  v1
Select : a 
t
0 ms  25 ms
Solve : a 
 8.333 sm2
3.0 s
or 8.3 m/s2 [backward]
Example with the Equations of
Motion
Calculate (b) the net force needed to cause that
acceleration if the mass of the car is 1200 kg.
Givens : a  8.333 sm2
m  1200 kg
Unknown: Fnet  ?
Example with the Equations of
Motion
Calculate (b) the net force needed to cause that
acceleration if the mass of the car is 1200 kg.
Givens : a  8.333 sm2
m  1200 kg
Unknown: Fnet  ?
Select : Fnet  ma


Solve : Fnet  1200 kg   8.333 sm2  10 000 N
or 10 000 N [backward]
More Practice
Newton’s 2nd Law of Motion Lab