# Homework Help: Challenge problem - Power

1. Feb 8, 2013

### jaumzaum

A force F is acting on an object whose mass is 1kg. The force in newtons is 4+4t², where t is the time. The angle in radians that the force does with the displacement is 2πt. If the object was initially at rest, estimate the power due to that force at t=3s.

I've tried to solve it but I've failed.
Here is what I got until now

P=power
v=velocity
m=mass
F=force
θ=angle force does with displacement
x=displacement
W=work

P = dW/dt = d($\vec{F}\bullet\vec{x})/dt = \vec{F} \bullet \vec{v} + d\vec{F}/dt\bullet\vec{x}$

W = $\int \vec{F}\bullet\ d \vec{x} = \int F.dx.cos(\theta)$
W = mv2/2

I don't know how to go on from here.

2. Feb 8, 2013

### TSny

The correct expression for the work done during an infinitesimal displacement is $dW = \vec{F}\cdot\vec{dx}$

So, what expression do you get for the power $P = dW/dt$?

Does the object move along a straight line? Or in 2 or 3 dimensions? [EDIT: Nevermind, I think it must be moving in 2 or 3 dimensions and F is the only force.]

Last edited: Feb 8, 2013
3. Feb 8, 2013

### jaumzaum

So the dF/dt . x goes out?

2 dimensions

4. Feb 8, 2013

Yes.

5. Feb 8, 2013

### I like Serena

Did you try to find an expression for the acceleration in the direction of the velocity?

6. Feb 8, 2013

### jaumzaum

Actually I've solved it considering the "displacement" term was instantaneous displacement (that has the direction of velocity). But I think displacement was used to design the vector that points from the initial position to the actual position of the object. In the first case:

a=Fcos(θ)/m = (4+4t²)cos(2πt)

Velocity is v = ∫a.dt = 6/π2
So P=F.v.cos(θ) = F.v = 240/π2

Is this right? What would be the result in the second case?

7. Feb 8, 2013

### I like Serena

Yes. That is right!

Displacement is the change in position in a (small) time interval.
It is always in the direction of the velocity.