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When is the acceleration
negative? positive?

What would happen if this bottle were dropped?

VIDEO

Newton's Masterpiece!

Sir
Isaac Newton 's own
first edition copy of his
Philosophiae Naturalis Principia Mathematica
with his handwritten corrections for the second edition.

The first edition was published under the imprint
of
Samuel Pepys who was
president of the
Royal Society . By the
time of the second edition, Newton himself had become president of the Royal
Society, as noted in his corrections.

The book can be seen in the
Wren Library of
Trinity College, Cambridge .

4. Standard Equation

F _{net}
=
ma
(N)
(Kg)(m/s ^{2} )

Newton - force needed to give a
1 kg of mass an acceleration
of 1 m/s ^{2}

Derived unit - A unit that comes from the combination of other units

N = Kg m/s^{2}

Fundamental unit
(opposite) m, s, Kg

Force Questions

Ex 1) What is the
net force necessary to give a 3.0 kg mass, initially at rest, an
acceleration of 5.0 m/s ^{2} ?

Ex 1) What is the
force necessary to give a
3.0 kg mass , initially at rest , an
acceleration of 5.0 m/s^{2} ?

F = ma

(3.0 kg)(5.0 m/s^{2} )
= 15 N

or

15 kgm/s^{2}

Ex 2) A net force of 30. nt accelerates an object by
15. m/s^{2} .

Find the objects mass. .

Ex 2) A force of
30. nt accelerates an object by
15. m/s^{2} .

Find the objects mass .

F = ma

30. N = m (15. m/s^{2} )

m = 2.0 kg

Ex 3) An unbalanced force
of 100. nt acts on a 50. kg mass for 5.0
seconds.

Find the acceleration of the mass.

Ex 3) An unbalanced force of
100. nt acts on a
50. kg
mass for 5.0 seconds .

Find the
acceleration
of the mass

F _{net}
= ma

100. nt = (50. kg) a

a = 2.0 m/s^{2}

Ex 4) A 10. kg cart moving with a
velocity of 2.0 m/s is brought to a stop in 2.0 sec.

Find magnitude
of the average force used to stop the cart.

Ex 4) A
10. kg
cart moving with a velocity of 2.0 m/s
is brought to a stop
in 2.0 sec .

Find
magnitude of the average force
used to the stop cart.

m = 10. kg

t = 2.0 sec

V_{i} = 2.0 m/s

V_{f} = 0 m/s

(stop) F = ?

F = ma

a = ?

a
=
V_{f}
- V_{i}
=
0 - 2.0 m/s
t
2.0 sec

a = -1.0 m/s^{2}

F _{net}
= ma

= (10. kg)(-1.0 m/s^{2} )

= -10. N

or

-10. kg • m/s^{2}

Ex 5) The object below has a mass of 5
kg.

Find its acceleration.

F_{net} = ma

40 N East = (5 kg)a

a = 8 m/s^{2 } East

Ex 6) The block above has an acceleration of 22. m/s^{2} .

What is the mass of the block?

5N^{2 } + 10N^{2 } = R^{2}

F_{net} = ma

F _{net} =11 N=m(22 m/s^{2} )

m = .50 kg

Ex 7) If the magnitude
of the block’s acceleration is 2.0 m/s^{2} ,
what is the mass
of the block?

F_{net} = ma

10 N Left = m(2 m/s^{2} )

m = 5 kg

What will the balloon do

when the
cart is pushed

to the left?

(Click on image)

Rutgers U

Balloon_in_car

The air in the container
tends to remain at rest. When the cart is pushed
to left, air pressure
builds up on the right
side of the container.

Try this at home!

VIDEO

Ranking
Task

Rank
the acceleration
from least to greatest.

Rank the accelerations
from
least to greatest

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Use any material on this site (w/ attribution)