
If you increase the speed of a 2.0kg air puck by
3.0 m/s in 4.0 s, what force do you exert on it?

Question Date: 20040303   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 accelel. in your example
is: a=3 m/s/4s= 0.75 m/s^2Thenfrom F=ma, the force
is F= (2 kg)*(0.75 m/s^2) = 1.5 kg*m/s^2 = 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 = 0.75Now use this in the 1st equation to
calculate the force F = M x a so...F = 2.0 x
0.75and the answer is, F = 1.5NForce 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 "deltav" 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 Fdt = mdv > F = mdv/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 kgm/sec^2 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 2.0kg 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/s^2. The force, therefore, would be 2.0
kg * 0.75 m/s^2 = 1.5kg*m/s^2, 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.





Copyright © 2017 The Regents of the University of California,
All Rights Reserved.
UCSB Terms of Use


