Why Was FNetx Changed to ma in the Physics Problem?

  • Thread starter Thread starter aprilrenee
  • Start date Start date
AI Thread Summary
The discussion centers on the application of Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration (FNet = ma). The user questions the transition from FNetx to ma in their physics problem, seeking clarification on this fundamental principle. The problem involves calculating the mass of a large telephone using a horizontal force and the resulting acceleration. The calculations provided demonstrate the relationship between force, mass, and acceleration, ultimately leading to the determination of the mass. Understanding this conversion is essential for solving physics problems involving forces and motion.
aprilrenee
Messages
2
Reaction score
0

Homework Statement


In 1988, a very large telephone was constructed in the NEtherlands. In order to move the phone, a 7980 N horizontal force was applied that result in an acceleration of 1.33 m/s^2 along level ground.


Homework Equations


I received this question on a practice test for my exam tomorrow. I also have the work (shown below.)

- I do not understand how/why FNetx was changed to ma.
Why was this done?

The Attempt at a Solution



FNety = Fw- W (0 because not moving vertically.)
FNetx = Fa-Fk
ma = Fa - (mu)mg
ma+ (mu)k mg = Fa
m (a+(mu)k g) = Fa
m = Fa / a+ (mu)k g
m = 7890 / [(1.33m/s^2) + (.180)(9.80m/s^2)]
m = 2579.18552 kg
m = 2580 kg

... So on and so forth.
 
Physics news on Phys.org
aprilrenee said:
- I do not understand how/why FNetx was changed to ma.
Why was this done?
Here's a hint. What is Isaac Newton's second law of motion?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Dynamics Skier on a Slope Problem
Replies
9
Views
5K
What are the forces acting on a child pulling a wagon up an inclined hill?
Replies
3
Views
4K
Calculating the Required Force to Prevent Sliding on an Incline Plane
Replies
3
Views
2K
Block and cart friction problem
Replies
6
Views
10K
Dynamics And Kinematics Problem
Replies
2
Views
2K
Calculating Distance: Free Body Diagram for Box Sliding with Friction
Replies
1
Views
2K
Find the magnitudes of all of the forces acting on the sled
Replies
5
Views
2K
Two masses connected by a weighted rope pulled vertically
Replies
5
Views
5K
Is Exponential Form Better for Balloon with Inserted Load Calculation?
4
Replies
170
Views
7K
Why does the speed change at the bottom of the loop-the-loop?
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top