Newton’s 2nd LawLab 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:

amFm

Fa

We usually write Neton’s 2nd law:

amF

m= 10 kgF2=300N F1=200N

+x

+y

Consider this situation, what is the acceleration of the box?

21010300200s

maakgNNamF

Consider this situation:1st is to draw a FBD for each mass

m2

m1

+

+

+

+

+m2: Tension, T

Weight, m2g

m1: Tension, T

Weight, m1g

2nd step is to write down Newton’s 2nd law for each mass is to draw a FBD for each mass:

m1: m1g – T = m1a Now we have a situation where we havem2: -m2g + T = m2a 2 equations and 2 unknowns. By adding +: m1g – m2g = m1a +m2a these equations we can solve for, a

21

21 )(

mm

gmma

Substituting back intoone of the other equationwe can solve for the tension

amgmT 22

Consider the situation when m1= 10 kg and m2 = 8 kg

Ns

m

s

mkgamgmT

s

m

kgkgsm

kgkg

mm

gmma

87)09.18.9)(8(

09.1810

8.9)810()(

2222

2

2

21

21

What if m1 = m2? Notice that a will equal 0, and the tension becomes equal to mg

Let’s look at this problem:If the object starts fromrest, how fast is it moving after 10 seconds?

10 kg

F = 400 N

20o

We need to find the object’s acceleration using Newton’s 2nd law! amF

2

0

6.37

)10(20cos400

s

ma

akgN

maFamF

x

x

xx

Now plug this acceleration into v = v0 + at

V = 0 + (37.6 m/s2) (10 s) = 376 m/s

mc

mw

Here is what we are doing in lab today:

mc

mw

T

Tmcg

FN

mwg

Write down the equations from Newton’s 2nd law:

Weight: mwg – T = mwaCart: T = mca Add the equations together

mwg = (mw + mc) a SOLVE FOR a : cw

w

mm

gma

mw

++

+

T

mc

mcg

FN

mwg

INCLINE PLANE

We use the same procedure, 1st FBD, 2nd we write newton’s second law:

Weight: mwg – T = mwaCart: T-mcsin = mca Add these guys: mwg-mcsin = mwa + mca

SOLVE FOR a:cw

cw

mm

gmma

)sin(

length

heigth1sin height