1. 22ndnd
LawLaw
The force exerted by anThe force exerted by an
object is equal to its massobject is equal to its mass
times its acceleration.times its acceleration.
F=maF=ma

2. 22ndnd
LawLaw

3. 5.2 The Second Law:
Force, Mass, and Acceleration
:
What is the mathematical
relationship between force,
mass, and acceleration?

4. 1. A 2kg object exerts a force of
10N. Calculate its acceleration.
F = 10 N or kg m/s2
m = 2 kg
a = ?
F = ma
Rearrange equation to find a=?
a = F = 10 kg m/s2
m 2
a = 5 m/s²

5. 2. An object with a mass of 2kg
has an acceleration of 5 m/s2
,
Calculate the force exerted by
the object.
m = 2 kg
a = 5 m/s²
F = ?
F = ma
No need to rearrange equation
F = 2kg x 5m/s²
= 10 N

6. 3. An object with an acceleration of
10 m/s2
exerts a force of 10 N.
Calculate its mass.
F = 10 N or (kg m/s2
)
a = 10 m/s²
m = ?
F = ma
Rearrange equation to find m =?
m = F = 10 N
a 10 m/s2
m = 1 kg

7. WE DOWE DO
1.1. What acceleration will resultWhat acceleration will result
when a 12 N net force iswhen a 12 N net force is
applied to a 3 kg object?applied to a 3 kg object?

8. ““We do”We do”
1. F = 12 N1. F = 12 N
m = 3 kgm = 3 kg
a = ?a = ?
F = maF = ma
Rearrange equation to find a =?Rearrange equation to find a =?
a =a = FF == 12 N12 N = 4 m/s= 4 m/s²²
m 3 kgm 3 kg

9. ““We do”We do”
2. A net force of 16 N causes2. A net force of 16 N causes
an object to accelerate at aan object to accelerate at a
rate of 5 m/srate of 5 m/s22
. Calculate the. Calculate the
mass of the object.mass of the object.

10. ““We do”We do”
2. F = 16 N2. F = 16 N
a = 5 m/sa = 5 m/s22
m = ?m = ?
F = maF = ma
Rearrange equation to find m =?Rearrange equation to find m =?
m =m = FF == 16 N16 N = 3.2 kg= 3.2 kg
a 5 m/sa 5 m/s22

11. ““We do”We do”
3. How much force is needed to3. How much force is needed to
accelerate a 66 kg skier at 1accelerate a 66 kg skier at 1
m/sm/s22
??

12. ““We do”We do”
3. F = ?3. F = ?
m = 66 kgm = 66 kg
a = 1 m/sa = 1 m/s22
F = ma Direct substitution:F = ma Direct substitution:
F = ma = 66kg x 1m/sF = ma = 66kg x 1m/s²²
= 66N= 66N

13. ““We do”We do”
4. What is the force on a 10004. What is the force on a 1000
kg elevator that is falling freelykg elevator that is falling freely
at 9.8 m/sat 9.8 m/s22
??

14. ““We do”We do”
4. F = ?4. F = ?
m = 1000 kgm = 1000 kg
a = 9.8 m/sa = 9.8 m/s22
F = maF = ma
Direct substitution:Direct substitution:
F = ma = 1000kg x 9.8m/sF = ma = 1000kg x 9.8m/s²²
= 9800 N= 9800 N

15. ““We do”We do”
5. What is the acceleration of a5. What is the acceleration of a
1400 kg car when a force of 28001400 kg car when a force of 2800
N is applied.N is applied.

16. ““We do”We do”
5. F = 2800 N5. F = 2800 N
m = 1400 kgm = 1400 kg
a = ?a = ?
F = maF = ma
Rearrange equation to find a =?Rearrange equation to find a =?
a =a = FF == 2800 N2800 N = 2 m/s= 2 m/s22
m 1400 kgm 1400 kg

17. Newton’s 2nd
Law proves that different masses accelerate to the
earth at the same rate, but with different forces.
We know that objects
with different masses
accelerate to the
ground at the same
rate.
However, because of
the 2nd
Law we know
that they don’t hit the
ground with the same
force.
F = maF = ma
98 N = 10 kg x 9.8 m/s/s98 N = 10 kg x 9.8 m/s/s
F = maF = ma
9.8 N = 1 kg x 9.89.8 N = 1 kg x 9.8
m/s/sm/s/s

18. Comparing the Mass of a dropped object
and the size of the impact crater
Amount of
sinkers
Mass of
Container (g)
Depth of
impact crater
(mm)
Diameter of
impact crater
(mm)
RANK of force
exerted (where
1=greatest force and
4 = least force
¼ full
½ full
¾ full
full

19. ““We Do” - ANSWERS:We Do” - ANSWERS:
 1. What acceleration will result when a 12 N net force applied to a 3 kg object?1. What acceleration will result when a 12 N net force applied to a 3 kg object?
a = F/m ; a =a = F/m ; a = 12 kg m/s/s12 kg m/s/s = 4 m/s/s= 4 m/s/s
3 kg3 kg
 2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s22
..
Determine the mass.Determine the mass.
m = F/a ; m =m = F/a ; m = 16 kg m/s/s16 kg m/s/s = 3.2 kg= 3.2 kg
5 m/s/s5 m/s/s
 3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec?3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec?
F = ma ; F =F = ma ; F = ((66 kg)1m/sec/sec or 66 N66 kg)1m/sec/sec or 66 N
 4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?
F = ma ; F =F = ma ; F = 10001000 kg x 9.8 m/sec/sec = 9800 Nkg x 9.8 m/sec/sec = 9800 N
 5. What is the acceleration of a 1400 kg car when a force of 2800 N is applied.5. What is the acceleration of a 1400 kg car when a force of 2800 N is applied.
a = F/m ; a =a = F/m ; a = 2800 kg m/s/s2800 kg m/s/s = 2 m/s/s= 2 m/s/s
1400 kg1400 kg