Why isn't the force in the direction of the velocity in the equation F=P/v?

[Mr Davis 97]

Homework Statement

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Homework Equations

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The Attempt at a Solution

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I simply want some guidance pertaining to if I did this problem correctly. Is the diagram correct? In addition, I have a specific question. We have the equation $\vec{F} = \frac{P}{\vec{v}}$. My question is if this equation is true, then why isn't the force in the direction of the velocity? I though that if a vector was written in terms of another vector, then the first vector had to be in the direction of the second (e.g. $\vec{v} = \frac{\mathrm{d} \vec{s}}{\mathrm{d} t}$, where velocity has to be in the direction of the displacement).

 

[Simon Bridge]

. We have the equation F⃗ =Pv⃗ \vec{F} = \frac{P}{\vec{v}}. My question is if this equation is true, then why isn't the force in the direction of the velocity?

The equation is not true.
You can derive the correct equation from the definitions of work and power.

 

[haruspex]

Homework Statement

See attached image

Homework Equations

See attached image

The Attempt at a Solution

See attached image

I simply want some guidance pertaining to if I did this problem correctly. Is the diagram correct? In addition, I have a specific question. We have the equation $\vec{F} = \frac{P}{\vec{v}}$. My question is if this equation is true, then why isn't the force in the direction of the velocity? I though that if a vector was written in terms of another vector, then the first vector had to be in the direction of the second (e.g. $\vec{v} = \frac{\mathrm{d} \vec{s}}{\mathrm{d} t}$, where velocity has to be in the direction of the displacement).

You can't divide by a vector. The equation is $\vec{F}.\vec{v} = {P}$. A dot product does not behave like scalar multiplication.

 

[NascentOxygen]

This is taxing my mind-reading finesse twice over!

It took me a while to figure out what the examiner was talking about exactly, and I'm still not 100% sure I have it right.

First, could you explain, in words, what you think the vector diagram should show?

Meanwhile, 160/4 ≠ 80

 

[haruspex]

Hint: write the F vector as the sum of its x and y components and substitute in the correct version of the power equation.

 

[Mr Davis 97]

This is taxing my mind-reading finesse twice over!

It took me a while to figure out what the examiner was talking about exactly, and I'm still not 100% sure I have it right.

First, could you explain, in words, what you think the vector diagram should show?

Meanwhile, 160/4 ≠ 80

The vector diagram should show the forces acting on the flat rock. There are three forces: the applied stick force, gravity, and the normal force. The normal and gravity cancel out, so the net force is that of stick, and since the question states that the stone is moving in the horizontal direction, it is specifically the x-component of the applied force that causes the acceleration. This is my reasoning. Anyhow, what exactly am I doing wrong? I'm not sure what to change because nobody has told me what I am doing incorrectly...

Also, you're right that 160/4 does not equal 80, but I evaluated the force again, correctly I think, and it still turned out to be 80 N.

 

[Simon Bridge]

what exactly am I doing wrong? I'm not sure what to change because nobody has told me what I am doing incorrectly...

I told you the equation you used was wrong and haruspex told you this was bacause you cannot divide by a vector... it is difficult to go into details more without doing the problem for you.

Basically:
You have not converted the vector equation to a magnitude equation correctly.
Check the maths definition of the dot product, or the physics definition of work and power.

 

[Mr Davis 97]

You have not converted the vector equation to a magnitude equation correctly.
You cannot divide by a vector... so what do you do instead. Check the maths definition of the dot product, or the physics definition of work and power.

So would this be correct?

$P = \frac {W}{t} = \vec{F}\cdot \vec{v} = Fv\cos \theta$
$F = \frac{P}{v\cos \theta} = \frac{160}{4.00\cos 60^{\circ}} = 80 N$

Then

$W = F_{x}d = (80)(2) \cos 60^{\circ} = 80 J$

Is this correct? If not, what am I doing wrong?

 

[Simon Bridge]

That's what I'd have done from the same information.

 

[NascentOxygen]

W=Fxd=(80)(2)cos60∘=80J
W = F_{x}d = (80)(2) \cos 60^{\circ} = 80 J

Your answer checks out. You'd calculate the same work if you said it's being moved at a steady speed for ½ second and her power is 160W.