Newton's Laws magnitude of the force

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Homework Help Overview

The discussion revolves around a physics problem involving Newton's laws, specifically focusing on the calculation of force magnitude and direction for an object experiencing a change in velocity over time. The object in question has a mass of 1.20 kg and undergoes a velocity change from 14.4 m/s to 28.8 m/s at an angle of 34 degrees to the x-axis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the breakdown of velocity into components and the application of momentum concepts to find force. There is an exploration of how to incorporate both x and y components of momentum into the calculations. Questions arise regarding the use of time in the calculations and the correct application of trigonometric functions to determine the direction of the force.

Discussion Status

Participants are actively engaging with the problem, offering various approaches to calculate the force and its direction. Some guidance has been provided regarding the relationship between momentum and force, and there is a recognition of the need to consider both components of momentum in the calculations. However, there is no explicit consensus on the final approach or solution yet.

Contextual Notes

Participants note the importance of time in the context of the problem, questioning its role in the calculations. There is also a discussion about the potential confusion arising from the relationship between momentum and force, particularly in terms of units and the application of trigonometric functions.

Mysteek
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Requesting guidance with this particular problem:

Homework Statement


A 1.20 kg object is moving in the x direction at 14.4 m/s. Just 3.31 s later, it is moving at 28.8 m/s at 34.0 degrees to the x axis.

a.) What is the magnitude of the force applied during this time?

b.) What is the direction of the force applied during this time?


Homework Equations


Fnet=ma?


The Attempt at a Solution


I first organized all my givens, that is:
mObject = 1.20kg
vObject = 14.4m/s
t = 3.31s
v2Object = 28.8m/s 34 degrees to x axis

I established a coordinate system so that positive would be ^ and -> , and negative would be the opposites. I then drew the corresponding vectors (14.4m/s and 28.8m/s at the angle).

Then I broke down the angled vector into its x and y components:

v2x = 28.8cos34 = 23.88m/s
v2y = 28.8sin34 = 16.1m/s

Here's where I get uncomfortable with my work:

I know Fnet = ma

So F = 1.2(v/t)
= 1.2(v2-v1 / t)
= 1.2kg((23.88m/s - 14.4m/s) / 3.31s)
= 3.44

I'm aware I completely neglected the y component, and I've been having trouble finding out where to fit it really. I'd love a push in the right direction if someone wouldn't mind - thanks!
 
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Hi mysteek, welcome to PF.
Force is defined rate of change of momentum.
Initial momentum Px is along the x direction. Resolve the final momentum into Px' and Py'.
Find the magnitude of change in momentum and hence the force.
 
Ah, momentum, brings back memories.

So then:

Px(initial) = mass of object * initial velocity
= 1.2kg(14.4m/s)
= 17.28

Px(final) = mass of object * x component final velocity
= 1.2kg(23.88m/s)
= 28.656

Py(final) = mass of object * y component final velocity
= 1.2kg(16.1m/s)
= 19.32

Now that I have the components, I can compute the change in momentum, thus providing me with force:

Fnet = Change in P

Change in Px = P(final) - P(initial)
= 28.656 - 17.28
= 11.376 N

Change in Py = 19.32 N

Magnitude of change in momentum would be:

Sqrt(11.376^2 + 19.32^2)
Sqrt(502.676)
= 22.4 N

Am I missing something? I didn't use the 3.31 seconds of time. Was ''time'' in the question just for the sake of throwing me off?
 
To get F divide change in momentum by time.
 
22.4 N would be the magnitude of the change of momentum, if I were to divide this by time I would get

22.4 / 3.31s

= 6.77 N

Oh, right! That makes perfect sense all the sudden, a bit more reading had me discover that momentum is in Newton seconds, so if I divided by time, seconds would cancel and I'd be left with Newtons, aka force.

At least I think my logic is correct. I'd like to think I understand the concept now.
 
Sorry for double posting - for some reason my browser isn't letting me click edit on my last post :(

When finding the direction of the force applied, could I use sine law?

Ex:

6.77N is the magnitude of the change of momentum, thus it would represent the hypotenuse of a right angle triangle, and because this vector is the hypotenuse, I can use:

(19.32 is the change in momentum in the y direction, aka opposite to the angle we need)

Sinx = opposite/hypotenuse
= 19.32 / 6.77

I get a math error when trying to find x. I assume I'm not using the correct numbers?

Crap, attempt number 2:

I think that the magnitude of the change in momentum has nothing to do with this particular section of the problem.

I found the hypotenuse earlier, did I not?

Sqrt(11.376^2 + 19.32^2)
Sqrt(502.676)
= 22.4 N s

This value is needed to keep units consistent, since the x and y components are also in N s.

Thus, I think I would use my same approach with this value?

sinx = o/h
= 19.32 / 22.4
x = 59.6 degrees?
 
Last edited:
tanθ = Py/Px
 

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