Solve Mechanics Problem: Find Magnitude of Force at t=2s

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Homework Help Overview

The problem involves a 3 kg object moving in a plane, with its position described by the equations x = 5t^2 - 1 and y = 3t^3 + 2. The goal is to find the magnitude of the force acting on the object at t = 2 seconds.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss finding expressions for velocity and acceleration through integration of position functions. There are questions about the relationship between position, velocity, and acceleration, as well as the correct mathematical operations to apply.

Discussion Status

Some participants have offered guidance on the relationships between position, velocity, and acceleration. There is an acknowledgment of confusion regarding the integration process, with one participant realizing the need to differentiate instead.

Contextual Notes

Participants are navigating the challenge of deriving acceleration from position functions and connecting these to the net force, while also grappling with the implications of their mathematical choices.

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1. The problem

A 3 kg object is moving in a plane with its x and y coordinates given by x = 5t^2 -1 and y = 3t^3 +2 where x and y are in meters and t is in seconds. Find the magnitude of the new force acting on this object at t= 2s

2. My attempt.

So my first attempt is to find a vector expression for acceleration. I integrated the expressions for x and y to give

V subx = 5/3 t^3 -t +c

To find the constant on integration I found the position of the particle at t=0 and t=2. Therefore calculating the velocity at 2 seconds. Thus the constant of integration for this expression is -4/3

The velocity at 2s being 10i+12j ms^-1

I have done the exact same process for velocity in the y direction and reached an expression for Vsuby .

My problem is I'm not sure how to get a valid expression for acceleration, and this seems like a really long winded way of answering this question.
 
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If you have an expression ##\mathbf{r}(t) = (x(t), y(t))## for position, wouldn't velocity be ##\dot{ \mathbf{r}}## and acceleration ##\ddot {\mathbf{r}}##?
 
HMPARTICLE said:
1. The problem

A 3 kg object is moving in a plane with its x and y coordinates given by x = 5t^2 -1 and y = 3t^3 +2 where x and y are in meters and t is in seconds. Find the magnitude of the new force acting on this object at t= 2s

2. My attempt.

So my first attempt is to find a vector expression for acceleration. I integrated the expressions for x and y to give

V subx = 5/3 t^3 -t +c

To find the constant on integration I found the position of the particle at t=0 and t=2. Therefore calculating the velocity at 2 seconds. Thus the constant of integration for this expression is -4/3

The velocity at 2s being 10i+12j ms^-1

I have done the exact same process for velocity in the y direction and reached an expression for Vsuby .

My problem is I'm not sure how to get a valid expression for acceleration, and this seems like a really long winded way of answering this question.

Why are you integrating? How are force and acceleration related? Define acceleration!
 
The net force on the object is the product of mass and acceleration. However I am being asked for the magnitude of the force on the object. I'm struggling to see how I can work with the expression for its displacement to get an expression for its acceleration so I can multiply its mass to get an expression for the net force
 
How is the position related to velocity? How is velocity related to acceleration?
 
ImageUploadedByPhysics Forums1406907739.713390.jpg


Got it! God I'm such a simpleton at times! I was inter grating when I should have been differentiating :(
 

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