Newton's Third law of motion problem.

AI Thread Summary
The discussion revolves around calculating the force of friction acting on an object moving across a table under two different applied forces. Given an applied force of 10 N resulting in an acceleration of 2 m/s², the mass is calculated to be 5 kg. In the second scenario, a 20 N force leads to an acceleration of 6 m/s², yielding a mass of 3.33 kg. The key point is that the applied force must be adjusted for friction to find the net force, which is essential for applying Newton's second law. The solution involves setting up equations based on the two scenarios to solve for the unknown friction force.
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Homework Statement


An object is free to move on a table, ex-
cept that there is a constant friction force f
that opposes the motion of the object when
it moves. If a force of 10 N pulls the object
across the table, the acceleration is 2 m/s2. If
a force of 20 N pulls the object across the ta-
ble, the acceleration is 6 m/s2. What is the
force of friction f?

a) 3.33 N
b) 1 N
c) 5 N
d) 10 N
e) none of these

Homework Equations



F= ma
and
F= -F'

and
F= μS where μ= coefficient of static friction

The Attempt at a Solution



i first find the mass in both case

F= 10 N and a= 2m/s/s
so
m= F/a = 10/2
m= 5kg

and again

F= 20N and a= 6m/s/s

so
m= F/a = 20/6
m= 3.33kg

what i do next..
or
either i must go to

F= μ S

please help me to solve it.. how i go further?
 
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abrowaqas said:
F= 10 N and a= 2m/s/s
so
m= F/a = 10/2
m= 5kg
10 N is the applied force, not the net force. To use Newton's 2nd law you need the net force.
 
then Doc Al

how will i find the net force..
will it be

Fnet = 20+10 = 30N ...
what i do next?
 
abrowaqas said:
how will i find the net force..
will it be

Fnet = 20+10 = 30N ...
No. In the first case, what two forces act? One is the applied force of 10 N. What's the other force?

You'll end up with an equation with 2 unknowns.

Then you'll get a second equation, using the second case (with the applied force of 20 N).

You'll solve those two equations for the two unknowns.
 
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