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

• Mimosapudica
In summary: 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...
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?
Relevant 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 ?

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?

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.

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!

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.

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..

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.

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 ^^

## 1. How is the power delivered when a block of mass m slides along calculated?

The power delivered when a block of mass m slides along is calculated by multiplying the force applied in the direction of motion by the velocity of the block. This can be written mathematically as P = F * v, where P is power, F is force, and v is velocity.

## 2. What is the unit of measurement for power?

The unit of measurement for power is watts (W), which is equivalent to joules per second (J/s). This unit is commonly used in the metric system to measure the rate at which work is done or energy is transferred.

## 3. How does the mass of the block affect the power delivered?

The mass of the block does not directly affect the power delivered. However, a heavier block may require more force to maintain a certain velocity, which would result in a higher power output. Additionally, the mass of the block may affect the acceleration and therefore the velocity, which would impact the power calculation.

## 4. Is the power delivered constant or does it change over time?

The power delivered can vary over time, depending on the force applied and the velocity of the block. If the force or velocity changes, the power delivered will also change. However, if the force and velocity remain constant, the power delivered will also remain constant.

## 5. Can the power delivered be negative?

Yes, the power delivered can be negative. This occurs when the force applied is in the opposite direction of the motion, causing the power to be negative. In this case, the block is either slowing down or moving in the opposite direction of the applied force, resulting in a negative power output.

Replies
4
Views
2K
Replies
41
Views
3K
Replies
31
Views
472
Replies
11
Views
2K
Replies
11
Views
445
Replies
2
Views
1K
Replies
18
Views
573
Replies
9
Views
3K
Replies
17
Views
2K
Replies
1
Views
828