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

#### Mimosapudica

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?
Homework Equations
P= force in the direction of velocity x velocity
Frictional force = coefficient of friction x normal reaction

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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#### kuruman

Homework Helper
Gold Member
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?

#### Mimosapudica

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?

#### Mimosapudica

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 cancelled ?!

#### kuruman

Homework Helper
Gold Member
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.
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.

#### Mimosapudica

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 cancelled 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?

#### kuruman

Homework Helper
Gold Member
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 cancelled ?!
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.

#### kuruman

Homework Helper
Gold Member
Normal reaction gets cancelled 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.

#### Mimosapudica

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?

#### kuruman

Homework Helper
Gold Member
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.
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.

#### Mimosapudica

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!!

#### kuruman

Homework Helper
Gold Member
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.

#### Mimosapudica

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..

#### kuruman

Homework Helper
Gold Member
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.

#### Mimosapudica

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 ^^

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

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