Calculating Power in a Force Problem with Changing Angle and Time

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A force F, defined as 4 + 4t² Newtons, acts on a 1kg object, with the angle of force relative to displacement given by 2πt radians. The discussion revolves around calculating the power at t=3s, starting from the initial rest state of the object. The correct approach involves using the relationship P = F·v·cos(θ), where velocity is derived from acceleration calculated as a = Fcos(θ)/m. The calculations lead to a power result of 240/π² at t=3s, confirming the methodology used. The conversation emphasizes understanding displacement as the change in position aligned with velocity direction.
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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.
 
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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.]
 
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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##?

So the dF/dt . x goes out?
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.]

2 dimensions
 
jaumzaum said:
So the dF/dt . x goes out?
Yes.
 
Did you try to find an expression for the acceleration in the direction of the velocity?
 
I like Serena said:
Did you try to find an expression for the acceleration in the direction of the velocity?

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?
 
Yes. That is right!

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