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If you increase the speed of a 2.0-kg air puck by
3.0 m/s in 4.0 s, what force do you exert on it?
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Question Date: 2004-03-03 | | Answer 1:
Well, force is mass times acceleration, and
acceleration is the change in velocity over (or
divided by) time.
So, we get:
a = 3.0/4.0 = 0.75 m2/s
and
F = m*a = 2.0*0.75 = 1.5 N
Pick some different values for mass, velocity, and
time and try the calculation yourself.
| | Answer 2:
The acceleration of the puck is the change in
velocity divided by the time interval needed to
change the velocity hence acceleration.
In your example is:
a = 3 m/s/4s= 0.75 m/s2
Then from F = m a, the force
is F= (2 kg)*(0.75 m/s2) = 1.5
kg*m/s2 = 1.5 Newtons.
| | Answer 3:
This is a simple problem for which you need to
use 2 different equations. Force = mass x
acceleration (F = M x a) and
acceleration = velocity (speed
in a particular direction) / time taken(a =
V/t).
First calculate the acceleration of the puck
using the second equation a = V/t so......
a = 3/4 = 0.75
Now use this in the 1st equation to
calculate the force F = M x a so...
F = 2.0 x 0.75 and the answer is,
F = 1.5N
Force is measured in Newtons. Best wishes.
| | Answer 4:
So, I'm glad you are studying physics! Let's
look at what this problem is asking:
"If you increase the speed of the puck BY 3.0
m/sec" - ok, so that means that the CHANGE in
speed is 3.0 m/sec. It does not matter what
the speed was before and after, just that the
CHANGE in speed is 3.0 m/sec.
We call this change "delta". So "delta-v" is 3
meters/second. That is our first clue. Now we take
this clue, and think about the problem - what is
it really all about? Well, something with a
mass is going at some initial velocity, and
something is done to it to make it change its
velocity, and that takes place over a certain
interval of time.
Aha - we say - this is a momentum conservation
type of question! The moving mass, in this
case an air puck, has an initial momentum, (mass x
velocity) before the problem "begins" and then
after the problem is over, it has some new
momentum (same mass x different velocity). We
know, from Newton's second law, that it
takes a force to change a body's momentum!
And that force has to act over a certain time
interval. So... we put this together in math
language:
F = force dt = time interval
m = mass
dv = change in velocity
F dt = m dv --> F = m dv/dt and that is our
final equation (REMEMBER TO ALWAYS SOLVE THE
ALGEBRA FIRST BEFORE YOU PLUG IN THE NUMBERS, NO
MATTER WHAT YOUR JUNIOR HIGH TEACHER DOES - IT IS
BEST TO ONLY INSERT THE ACTUAL NUMBERS IN THE
FINAL STEP!!!)
NOW we can plug in the numbers:
From the given s, m = 2.0 kg, dt = 4.0seconds, dv
= 3.0 m/sec
F = (2.0 kg)(3.0 m/s) / (4.0 sec) =
1.5 kg-m/sec2 or 1.5 Newtons.
Hope this helps. It is important, even in the
beginning stages of learning physics, to use
GOOD PROBLEM SOLVING strategies!
| | Answer 5:
If you increase the speed of a 2.0-kg air puck
by 3.0 m/s in 4.0 s, what force do you exert on
it? This sounds a lot like a homework
question so I'm not going to give the exact answer
but, as you know, Newton's second law says
Force = Mass X Acceleration. In addition, if
you assume constant acceleration, Acceleration
= Change in Velocity / Time of Change.
You should be able to get the answer from this... | | Answer 6:
This is perhaps the most straightforward
physics question that you could ask. According to
Newton's second law, force is defined as the
mass of an object times its acceleration,
which can be expressed mathematically as
F = ma.
Acceleration is
simply defined as the change in velocity over
time.
The mass that you have given is 2.0 kg. The
change in velocity is 3.0 m/s, and the time in
which said acceleration took place is 4 s.
Therefore,
the acceleration is
(3.0 m/s)/(4.0 s) = 0.75 m/s2.
The force, therefore,
would be
2.0 kg * 0.75 m/s2 =
1.5kg*m/s2,
or 1.5 Newtons of
force (Because Issac Newton defined "force",
the standard international unit of force has been
named after him.
9.8 Newtons is about 2.2 pounds, 1.5 Newtons
is a little bit less than half a pound. Click Here to return to the search form.
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