Power, force, velocity and tension...

[hello478]

Homework Statement

in finding the speed, do we use the tension force or the weight?
and what is f, wouldnt it be zero? as resultant force is zero
what does f in this equation mean?

Relevant Equations

p=fv

my attempt: i solved it all correct but i dont understand a few things mentioned above...
82.04 * v = 56
so i got v as 0.68 m/s which is correct
but i dont understand the concept...

 

[BvU]

Does this also have a history that I can't see ?

my attempt: i solved it all correct but i dont understand a few things mentioned above...

Namely ?

$\$

 

[hello478]

Does this also have a history that I can't see ?



Namely ?

$\$

i did
82.04 * v = 56
so i got v as 0.68 m/s which is correct
but i dont understand the concept...

 

[BvU]

82.04 ?

 

[hello478]

82.04 ?

430 N *sin 11 for the component of weight along slope...

 

[BvU]

Yes, I reverse-engineered that (reverse engineering is second best after telepathy for a homework helper )

So your question is conceptual. Like: why $P = \vec F \cdot \vec v$ ?

With constant speed the fet force is zero. Gravity one way, tension the other. All work done is converted to potential energy.
Work is force times distance, power is work per time. So power is force times distance per time is distance times speed.

That what you meant ?

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[hello478]

Yes, I reverse-engineered that (reverse engineering is second best after telepathy for a homework helper )

So your question is conceptual. Like: why $P = \vec F \cdot \vec v$ ?

With constant speed the fet force is zero. Gravity one way, tension the other. All work done is converted to potential energy.
Work is force times distance, power is work per time. So power is force times distance per time is distance times speed.

That what you meant ?

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i understood what you said
and im really sorry for not asking properly
but i meant

in finding the speed from the formula p=fv , do we use the tension force or the weight? because they both have same magnitude so i didnt understand...

and what is f, wouldnt it be zero? as resultant force is zero due to constant speed

what does f in this equation mean? i know its the force but is it the total force? or resultant? or smth else?

 

[BvU]

i understood what you said
and im really sorry for not asking properly

No need to apologize -- we help voluntarily and you cooperate adequately

but i meant

in finding the speed from the formula p=fv , do we use the tension force or the weight? because they both have same magnitude so i didnt understand...

and what is f, wouldnt it be zero? as resultant force is zero due to constant speed

what does f in this equation mean? i know its the force but is it the total force? or resultant? or smth else?

I subtly wrote $P = \vec F \cdot \vec v$ to emphasize the vector character. $\vec F$ and $\vec v$ pointing in the same direction yields a positive power: the power that the motor delivers.

There is also the component along the slope of the weight, which points in opposite direction wrt $\vec v$. So ' gravity does negative work ', (i.e. work is done against gravity) to the tune of $-P$. And energy is conserved !

Time for my dinner break ...


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[hello478]

Time for my dinner break ...

please help me when you are free

 

[erobz]

Use the net force acting on the body. Disregard

The work expended by the motor is transferred to the body though the tension in the rope. Nice save @MatinSAR

 

[MatinSAR]

$$dW=\vec F . d\vec r = \vec F . \vec v dt$$ Power is defined as $dW/dt$. In above equation divide both sides by $dt$. You get: $$\dfrac {dW}{dt}=\vec F . \vec v$$ In this equation $P$ is power of force $\vec F$. In your question you have power of the motor(tension force). When you want to use $P=\vec F . \vec v$ for power of motor you should only consider the force that acts on moving object by the motor. So $\vec F$ is only tension force. Using N's 2nd law you can find it's magnitude and direction in term of gravitional force.

 

[hello478]

Using N's 2nd law you can find it's magnitude and direction in term of gravitional force.

i dont get this part...

 

[MatinSAR]

i dont get this part...

Two forces are acting on the object. Tension force and gravitional force.
$$\vec T + m \vec g = m \vec a=0$$
Can't you find tension force? Then how did you solve the question!

 

[hello478]

Two forces are acting on the object. Tension force and gravitional force.
$$\vec T + m \vec g = m \vec a=0$$
Can't you find tension force? Then how did you solve the question!

i used t=-mg for when a=0
not the 2nd law of motion...

 

[MatinSAR]

i used t=-mg for when a=0
not the 2nd law of motion...

Wrong. This way tension force should be 430.But you wrote 82.04.

 

[MatinSAR]

You need to find tension force using object's weight. The only way is to use Newton's 2nd law for $\vec a=0$.

 

[hello478]

You need to find tension force using object's weight. The only way is to use Newton's 2nd law for $\vec a=0$.

no like, but the component of the weight along the slope would be equal to the tension but in opposite direction as no other force is acting in that direction... where is the newtons second law?

 

[MatinSAR]

no like, but the component of the weight along the slope would be equal to the tension but in opposite direction as no other force is acting in that direction... where is the newtons second law?

This is correct. There is no acceleration along the slope so according to N's 2nd law there is no net force. This means that $T - mg\sin \theta$ should be 0.

@erobz Thanks for pointing out the mistake.

 

[hello478]

This is correct. There is no acceleration along the slope so according to N's law there is no net force. This means that $T + mg\sin \theta$ should be 0.

yes, so then i would use the tension force and multiply with velocity, because it is the force causing upward motion? is this explanation correct?

 

[erobz]

This is correct. There is no acceleration along the slope so according to N's 2nd law there is no net force. This means that $T + mg\sin \theta$ should be 0.

If you aren’t using the vector notation you should probably make one of those terms negative?

 

[hello478]

If you aren’t using the vector notation you should probably make one of those terms negative?

mg, would be negative as it opposes tension? and its downwards..

 

[MatinSAR]

yes, so then i would use the tension force and multiply with velocity, because it is the force causing upward motion? is this explanation correct?

We don't care about whaat causes the motion.

You want to find speed of the object using $P=\vec F. \vec v$ and you have power of the Tension force so that F in above equation represents tension force.

 

[hello478]

We don't care about whaat causes the motion.

You want to find speed of the object using $P=\vec F. \vec v$ and you have power of the Tension force so that F in above equation represents tension force.

i dont understand this... please explain more...

 

[MatinSAR]

If you aren’t using the vector notation you should probably make one of those terms negative?

Yes. I am using latex on phone currently and it's a bit complicated for me. Thanks.

 

[MatinSAR]

Let's start from the beginning. Is this post clear @hello478 ? Do you understand why you should only use tension force to find the speed?

$$dW=\vec F . d\vec r = \vec F . \vec v dt$$ Power is defined as $dW/dt$. In above equation divide both sides by $dt$. You get: $$\dfrac {dW}{dt}=\vec F . \vec v$$ In this equation $P$ is power of force $\vec F$. In your question you have power of the motor(tension force). When you want to use $P=\vec F . \vec v$ for power of motor you should only consider the force that acts on moving object by the motor. So $\vec F$ is only tension force.

 

[haruspex]

i dont understand this... please explain more...

The tension force exists because the motor is pulling with force F. We can look at it from the point of view of the wire or of the motor.
The motor, presumably, is winding the wire onto a spool. The force it exerts is $\vec F$, positive in the downslope direction. The velocity with which this is winding wire onto the spool is $\vec v$, also positive in the downslope direction. The power is $\vec F\cdot \vec v$, also positive.
From the wire's perspective, the force is positive upslope, but stationary, so the wire is not producing any power.

 

[MatinSAR]

The tension force exists because the motor is pulling with force F. We can look at it from the point of view of the wire or of the motor.
The motor, presumably, is winding the wire onto a spool. The force it exerts is $\vec F$, positive in the downslope direction. The velocity with which this is winding wire onto the spool is $\vec v$, also positive in the downslope direction. The power is $\vec F\cdot \vec v$, also positive.
From the wire's perspective, the force is positive upslope, but stationary, so the wire is not producing any power.

Is it incorrect to assume that the motor, which is pulling object up the slope, is exerting a tension force, $F$ which is equal to $m g \sin \theta$? Given that the power of the motor is provided, we can calculate the speed using the formula $v = P/F$ easily. Is this way wrong in your opinion?

 

[haruspex]

Is it incorrect to assume that the motor, which is pulling object up the slope, is exerting a tension force, $F$ which is equal to $m g \sin \theta$? Given that the power of the motor is provided, we can calculate the speed using the formula $v = P/F$ easily. Is this way wrong in your opinion?

I was just trying to be careful. The force exerted by the motor on the wire acts down the slope, but the block is moving up the slope, so you might calculate the power as negative.

 

[BvU]

Two forces are acting on the object. Tension force and gravitional force.
$$\vec T + m \vec g = m \vec a=0$$
Can't you find tension force? Then how did you solve the question!

I definitely see three forces and write$$\vec T +m\vec g +\vec N = m\vec a = \vec 0$$


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[MatinSAR]

I definitely see three forces and write$$\vec T +m\vec g +\vec N = m\vec a = \vec 0$$


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And I definitely won't use latex on my phone again. I always forget a part of the answer while using it on my phone. Thank you for pointing out the mistake.