Newton’s 2nd Law
Lab 4
Only 8 more to go!!
Newton’s 2nd Law says that the acceleration of an object is directly
proportional to the total force applied and inversely proportional to the
object’s mass. Mathematically:

 F
a
m
We usually write Neton’s 2nd law:


 F  ma


 F  ma
Consider this situation, what is the acceleration of the box?
+y
F2=300N
F1=200N
m= 10 kg
+x




m
 F  ma 200 N  300 N  10kg a  a  10 s 2
Consider this situation:
1st is to draw a FBD for each mass
m 1:
Tension, T
m2 :
Weight, m1g
Tension, T
Weight, m2g
+
+
+
m2
+
+
m1
2nd step is to write down Newton’s 2nd law for each mass is to draw a FBD
for each mass:
m 1:
m2:
+:
m 1g – T = m 1 a
-m2g + T = m2a
m1g – m2g = m1a +m2a
(m1  m2 ) g
a
m1  m2
Now we have a situation where we have
2 equations and 2 unknowns. By adding
these equations we can solve for, a
Substituting back into
one of the other equation
we can solve for the tension
T  m2 g  m2a
Consider the situation when m1= 10 kg and m2 = 8 kg
m
(10kg  8kg)9.8 2
(m1  m2 ) g
m
s
a

 1.09 2
m1  m2
10kg  8kg
s
m
m
T  m2 g  m2 a  (8kg)(9.8 2  1.09 2 )  87 N
s
s
What if m1 = m2?
becomes equal to mg
Notice that a will equal 0, and the tension
Let’s look at this problem:
If the object starts from
rest, how fast is it moving
after 10 seconds?
F = 400 N
20o
10 kg
We need to find the object’s acceleration
using Newton’s 2nd law!


 F  ma


 F  ma  Fx  max
400 N cos 20  (10kg) a x
0
m
a x  37.6 2
s
Now plug this acceleration into v = v0 + at
V = 0 + (37.6 m/s2) (10 s) = 376 m/s
Here is what we are doing in lab today:
mc
FN
mc
T
T
mcg
mw
mw
Write down the equations from Newton’s 2nd law:
Weight:
Cart:
m wg – T = m wa
T = mca
Add the equations together
mwg = (mw + mc) a
SOLVE FOR a :
mw g
a
mw  mc
m wg
+
INCLINE PLANE
FN
 heigth 

  sin 
 length 
1
+
mc

mcg

T
+
mw
height
m wg
We use the same procedure, 1st FBD, 2nd we write newton’s second law:
Weight:
Cart:
SOLVE FOR a:
mwg – T = mwa
T-mcsin = mca Add these guys: mwg-mcsin = mwa + mca
(mw  mc sin  ) g
a
mw  mc