Calculating Power in a Force Problem with Changing Angle and Time

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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([itex]\vec{F}\bullet\vec{x})/dt = \vec{F} \bullet \vec{v} + d\vec{F}/dt\bullet\vec{x}[/itex]


W = [itex]\int \vec{F}\bullet\ d \vec{x} = \int F.dx.cos(\theta)[/itex]
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.]
 
Last edited:
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
 
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?