Calculating Power in a Force Problem with Changing Angle and Time

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Homework Help Overview

The problem involves calculating the power exerted by a force acting on a 1 kg object, where the force is defined as 4 + 4t² Newtons and the angle between the force and displacement changes with time as 2πt radians. The original poster seeks to estimate the power at t=3s, having attempted to derive relevant equations but feeling uncertain about the next steps.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expression for power as the derivative of work with respect to time, questioning the dimensionality of the object's motion and the implications for the force's application. Some explore the relationship between force, velocity, and angle, while others consider the instantaneous nature of displacement in their calculations.

Discussion Status

Participants have provided clarifications on the expressions for work and power, with some confirming the validity of the original poster's approach. There is an ongoing exploration of different interpretations of displacement and its relationship to velocity, indicating a productive dialogue without a definitive consensus.

Contextual Notes

Some participants note the potential complexity of the problem due to the changing angle and force over time, as well as the initial condition of the object being at rest. The discussion reflects uncertainty regarding the dimensionality of motion and the definitions of terms used in the context of the problem.

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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
 
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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