What are the equations for calculating work and energy in a physics problem?

[Slnj325]

Question 1:

Homework Statement

Pinkfield slides a 1000 kg piano at a constant speed against a frictional force of 5000 N through a distance of 2.0m

Homework Equations

W = Fa x d?

The Attempt at a Solution

a = 5000/1000 = 5m/s^2

I tried to use this formula i found somewhere :

Ei + Wf = Ef

Ei = W

Wf = friction x distance

Ef = 0.5(1000)V^2

0 (2) - 5000(2) = 0.5(1000)V^2

= -10000 = 500 V^2

V^2 = 10000+ 500
V = Squareroot (10500)
V = 102.5m/s

W = 1/2(m)V^2
W = 1/2(1000)102.5^2

I stopped here because i knew i was doing something wrong.

Is their away to find Fa?

Question 2 :

Homework Statement

A 100 g glider is moved up a frictionless air track at a constant velocity. The length of the track is 1.0 m and the height of the elevated end of the track is 30 cm.

Homework Equations

W = mg x h

The Attempt at a Solution

W = 0.1kg(9.8)(0.3)
W = 0.294 Nm

Is this Right?

 

[berkeman]

Question 1:

Homework Statement

Pinkfield slides a 1000 kg piano at a constant speed against a frictional force of 5000 N through a distance of 2.0m

Homework Equations

W = Fa x d?

The Attempt at a Solution

a = 5000/1000 = 5m/s^2

I tried to use this formula i found somewhere :

Ei + Wf = Ef

Ei = W

Wf = friction x distance

Ef = 0.5(1000)V^2

0 (2) - 5000(2) = 0.5(1000)V^2

= -10000 = 500 V^2

V^2 = 10000+ 500
V = Squareroot (10500)
V = 102.5m/s

W = 1/2(m)V^2
W = 1/2(1000)102.5^2

I stopped here because i knew i was doing something wrong.

Question 2 :

Homework Statement

A 100 g glider is moved up a frictionless air track at a constant velocity. The length of the track is 1.0 m and the height of the elevated end of the track is 30 cm.

Homework Equations

W = mg x h

The Attempt at a Solution

W = 0.1kg(9.8)(0.3)
W = 0.294 Nm

Is this Right?

On #1, you are given the frictional force that opposes you, and how far you push it with constant speed. That is all you need to calculate the work done. You do not need the mass or the coefficient of friction.

Your #2 looks correct.

 

[Slnj325]

On #1, you are given the frictional force that opposes you, and how far you push it with constant speed. That is all you need to calculate the work done. You do not need the mass or the coefficient of friction.

Your #2 looks correct.

So for # 1 it will be :

W = 5000N x 2.0m = 10000Nm?

so 10000Nm is the Work?

 

[berkeman]

So for # 1 it will be :

W = 5000N x 2.0m = 10000Nm?

so 10000Nm is the Work?

Yep. Sometimes questions will include info (like the mass in this case) that is not needed to solve the problem. That's part of checking to see if you understand the fundamental underlying mechanisms. In this case they openly gave you the resisting force due to friction, instead of having you calculate it from the mass and mu of the piano.

 

[berkeman]

BTW, who is Pinkfield?

 

[Slnj325]

Yep. Sometimes questions will include info (like the mass in this case) that is not needed to solve the problem. That's part of checking to see if you understand the fundamental underlying mechanisms. In this case they openly gave you the resisting force due to friction, instead of having you calculate it from the mass and mu of the piano.

Oh Ok I see Thanks a Lot For Your Help :)

BTW, who is Pinkfield?

Pinkfield was the name of the Person pushing the Piano :/

 

[berkeman]

Pinkfield was the name of the Person pushing the Piano :/

Well, yeah, I kind of figured that part out. Just was wondering if there was some story behind the name. No Biggie.

 

[Slnj325]

Well, yeah, I kind of figured that part out. Just was wondering if there was some story behind the name. No Biggie.

Oh no It was just some Random Question from a Handout which looks like it's been taken out of old Text book or something.

 

[collinsmark]

Hello Slnj325,

Welcome to Physics Forums!

The rules are only one homework problem per thread. But I'll comment anyway.

Question 1:

Homework Statement

Pinkfield slides a 1000 kg piano at a constant speed against a frictional force of 5000 N through a distance of 2.0m

Homework Equations

W = Fa x d?

You have the wrong formula for work. For a uniform (constant) force,

W = F·d.​

In words, work is the vector dot product of force and displacement. That works out to be the same thing as

W = Fdcosθ,​

where θ is the angle between the force vector and the displacement vector. (In the above formula, the unbolded F and d are the magnitudes of the vectors.)

The Attempt at a Solution

[...]

I stopped here because i knew i was doing something wrong.

You haven't actually mentioned what the question really is. I assume you are supposed to find the work done on friction. Is that correct?

There's no point in solving for the acceleration. You already know what it is anyway! What is the acceleration of anything moving at a constant velocity?

Question 2 :

Homework Statement

A 100 g glider is moved up a frictionless air track at a constant velocity. The length of the track is 1.0 m and the height of the elevated end of the track is 30 cm.

Homework Equations

W = mg x h

Be careful about using the 'x' symbol here. It means something very specific when dealing with vectors, and that's different from what you want it to mean here.

The Attempt at a Solution

W = 0.1kg(9.8)(0.3)
W = 0.294 Nm

Is this Right?

Again, you haven't really told us what the actual question was. But if you are trying to find the work done on gravity, then yes, your final answer is right.

But before you move on, I ask you to take a step back and contemplate how you got that answer. Do you see how your answer is really just,

W = F·d = Fdcosθ ?

(Remember, θ is not necessarily the angle of any incline. It is the angle between the force and displacement vectors. [and F is the force of gravity, and d is the displacement of the glider])

[Edit: I see berkeman beat me to the punch a few times over. I suppose I should type much faster. ]

 

[Slnj325]

Hello Slnj325,

Welcome to Physics Forums!

The rules are only one homework problem per thread. But I'll comment anyway.

You have the wrong formula for work. For a uniform (constant) force,

W = F·d.​

In words, work is the vector dot product of force and displacement. That works out to be the same thing as

W = Fdcosθ,​

where θ is the angle between the force vector and the displacement vector. (In the above formula, the unbolded F and d are the magnitudes of the vectors.)

You haven't actually mentioned what the question really is. I assume you are supposed to find the work done on friction. Is that correct?

There's no point in solving for the acceleration. You already know what it is anyway! What is the acceleration of anything moving at a constant velocity?

Be careful about using the 'x' symbol here. It means something very specific when dealing with vectors, and that's different from what you want it to mean here.

Again, you haven't really told us what the actual question was. But if you are trying to find the work done on gravity, then yes, your final answer is right.

But before you move on, I ask you to take a step back and contemplate how you got that answer. Do you see how your answer is really just,

W = F·d = Fdcosθ ?

(Remember, θ is not necessarily the angle of any incline. It is the angle between the force and displacement vectors.)

[Edit: I see berkeman beat me to the punch a few times over. I suppose I should type much faster. ]

Yes i was trying to Find the Work for these Two Question i forgot to mention that.

For Question 1 the person is pushing straight and not in an angle so i don't know how cosθ i used in the equation :? I haven't Really learned much about this my teacher taught my class about pushing objects in a angle and finding it's work but i didn't really understand it.

But the Answer is W = 10 000 Nm using W = 5000N* 2.0m?

 

[collinsmark]

For Question 1 the person is pushing straight and not in an angle so i don't know how cosθ i used in the equation :?

It technically still fits in. It's just that it's trivial if both the force and displacement are parallel (i.e. it's really easy if the angle between the force and displacement vectors is zero). What's the cos0?

But the Answer is W = 10 000 Nm using W = 5000N* 2.0m?

Yes, that's right

(5000 [N])(2 [m])cos0 = 10 000 [J].​