Calculating the power delivered when a block of mass m slides along

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The discussion revolves around calculating the power delivered by a man pulling a block with mass m while considering friction. The net force acting on the block is derived from the tension T and the frictional force, but the power delivered is primarily dependent on the tension and the velocity, expressed as P = Tv cos(theta). Friction does oppose the motion but does not directly factor into the power equation since the power is defined by the force exerted by the man and the resulting velocity. The normal reaction cancels with the vertical component of the tension, leading to the conclusion that while friction exists, it does not alter the power calculation directly. Ultimately, the power delivered by the man is independent of other forces acting on the block, emphasizing the importance of understanding the relationship between force, velocity, and power.
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Homework Statement
A block of mass m slides along the track with kinetic friction μ. A man pulls the block through a rope which makes an angle theta with the horizontal. The block moves with a constant speed v. Power delivered by the man is?
Relevant Equations
P= force in the direction of velocity x velocity
Frictional force = coefficient of friction x normal reaction
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Force along the horizontal would be
T cos(theta)
Frictional force (which is in the opposite direction )= μmg
So net force in the direction of velocity = Tcos(theta)-μmg
P= [Tcos(theta)-μmg]v
But this is not so, the right answer is given to be Tvcos(theta). Why should we not consider the frictional force acting here ?
 
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In this case the magnitude of the frictional force is not ##\mu mg##. What is it? Is the power delivered by the man related to the net force?
 
kuruman said:
In this case the magnitude of the frictional force is not ##\mu mg##. What is it?

Isn’t frictional force equal to the normal reaction x coefficient of friction?
Otherwise μk= tan (alpha) but this angle is the angle between the force and the normal reaction, which would be (90+ Theta)?

I’m confused because if there is a coefficient of friction for the surface then friction must act on it while the block moves, right?
 
kuruman said:
Is the power delivered by the man related to the net force?

Well yes, but only the component (horizontal)of the force in the direction of the velocity. The other component (vertical)cancels with the normal reaction.. does that mean there won’t be any friction if the normal reaction force gets canceled ?!
 
Mimosapudica said:
Isn’t frictional force equal to the normal reaction x coefficient of friction?
That it is. What is the normal reaction? Look at your drawing and write an equation for the balance of forces in the vertical direction.
Mimosapudica said:
Otherwise μk= tan (alpha) but this angle is the angle between the force and the normal reaction, which would be (90+ Theta)?
I’m confused because if there is a coefficient of friction for the surface then friction must act on it while the block moves, right?
Friction does act on the block while it moves. You don't need to find an expression for μ. Assume that it is given.
 
kuruman said:
That it is. What is the normal reaction? Look at your drawing and write an equation for the balance of forces in the vertical direction.

Normal reaction gets canceled by the vertical component of T. (Tsin(theta))

Sorry, but if friction was provided, it would still be opposing the motion of the block, so won’t it reduce the net force acting in the direction of velocity?
 
Mimosapudica said:
Well yes, but only the component (horizontal)of the force in the direction of the velocity. The other component (vertical)cancels with the normal reaction.. does that mean there won’t be any friction if the normal reaction force gets canceled ?!
Set friction aside and focus on the basics. You are looking for the power delivered by the man. What is the definition of power (I am not looking for Fv) delivered by a force? Note that the force exerted by the man is the tension T.
 
Mimosapudica said:
Normal reaction gets canceled by the vertical component of T. (Tsin(theta))

Sorry, but if friction was provided, it would still be opposing the motion of the block, so won’t it reduce the net force acting in the direction of velocity?
Yes to all. See post #7.
 
kuruman said:
Set friction aside and focus on the basics. You are looking for the power delivered by the man. What is the definition of power (I am not looking for Fv) delivered by a force? Note that the force exerted by the man is the tension T.
Ohh.. so power purely depends on the man ie the force he provides and the resulting velocity but not on any of the other forces that might act here. Am I right ?
Just wondering tho, will work done by the man be dependent on the friction? Will we have to add the frictional force in that case?
 
  • #10
Mimosapudica said:
Ohh.. so power purely depends on the man ie the force he provides and the resulting velocity but not on any of the other forces that might act here. Am I right ?
You are right.
Mimosapudica said:
Just wondering tho, will work done by the man be dependent on the friction? Will we have to add the frictional force in that case?
If you write the definition of power as I suggested in post #7, and apply it, you will answer that question on your own.
 
  • #11
kuruman said:
You are right.

If you write the definition of power as I suggested in post #7, and apply it, you will answer that question on your own.

Ohhh right .. P= fv= W/t .. so W=fvt
So again friction isn’t playing a part.
Thank you! 😊
 
  • #12
Mimosapudica said:
Ohhh right .. P= fv= W/t .. so W=fvt
So again friction isn’t playing a part.
Thank you! 😊
Not quite. For a constant force,$$P=\frac{dW}{dt}=\frac{d}{dt}(\vec F\cdot \vec s)=\vec F\cdot\frac{d\vec s}{dt}=\vec F \cdot \vec v$$The vectors and the dot product make a difference which is one of the important points in this problem.
 
  • #13
kuruman said:
Not quite. For a constant force,$$P=\frac{dW}{dt}=\frac{d}{dt}(\vec F\cdot \vec s)=\vec F\cdot\frac{d\vec s}{dt}=\vec F \cdot \vec v$$The vectors and the dot product make a difference which is one of the important points in this problem.

Ok so that’s why the force in the power equation is dependent on the man alone right ? Since it was derived from the work..
 
  • #14
Right. Each of the forces acting on the block is associated with its own power. The man delivers power while the friction dissipates all of it because the net force does no work since it is zero.
 
  • #15
kuruman said:
Right. Each of the forces acting on the block is associated with its own power. The man delivers power while the friction dissipates all of it because the net force does no work since it is zero.

I see! Thanks again for clearing this concept ^^
 
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