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

  • Thread starter Thread starter HMPARTICLE
  • Start date Start date
  • Tags Tags
    Mechanics
HMPARTICLE
Messages
95
Reaction score
0
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.
 
Physics news on Phys.org
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 :(
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top