Friction and W=delta KE

[G1SG2]

A box of mass 5 kg slides on a horizontal surface with initial speed 6 m/s. It feels no forces except that of gravity and the force from the surface. If the coefficient of kinetic friction between the box and the surface is 0.4, how much work is done by friction in bringing this box to rest?

The answer is -90 J. I solved this by finding the acceleration and then found the distance traveled through a kinematic equation. Then I used w=F*D to get the answer. However, the TPR explanation is w=delta KE, and that you could just plug in the numbers in 1/2mv^2 to get 90 J. I thought we couldn't use that equation when friction is involved? Or am I mistaken?

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[Seraph 84]

Net work done on an object is equal to its change in kinetic energy. The equation can be used for just about anything regardless of the number of forces involved.

 

[Doodl3s]

Yea man... think about it. If it goes from full kinetic energy to a complete stop, and friction is the only force acting on it. The work done by friction MUST be the lost KE, which in this case, is the total KE

 

[matth87]

The formula for work = change in KE is derived from BIG 5 #5

v^2 = v^2 (initial) + 2ad

plug F/m = a

F x d = W

Solve for work and you get (delta) KE.

So the kinematic equations do work because that's where the W = (delta)KE is derived from.

 

[sv3]

Net work done on an object is equal to its change in kinetic energy. The equation can be used for just about anything regardless of the number of forces involved.

this is obvious but just to be safe: the above applies when the object has only KE. So if it has any other type of energy as well, can't apply work energy theorm. Perhaps TME or something like that.

 

[G1SG2]

Okay, thanks everyone.

 

[Seraph 84]

this is obvious but just to be safe: the above applies when the object has only KE. So if it has any other type of energy as well, can't apply work energy theorm. Perhaps TME or something like that.

It can have potential energy as well.

If I'm holding a box steady at 1m and raise it to 2m, the net work done on the object is zero because it experienced zero change in it's kinetic energy even though it gained potential energy.

 

[minutemen11]

are you saying that a weightlifter does no net work lifting? PE is in joules which is work i thought?
Also how would you write out an expression for a lifter holding the weight above his head? There cant be 0 work done at this point right?

 

[Seraph 84]

Precisely! A muscle man that squats 300kg is doing zero net work on the weight. Tell him that next time you're at the gym
But don't make him angry.....you won't like him when he's angry.....

If he's just holding the weight above his head, then he is doing no work on the weight. The total work that the lifter does on the weight will be equal to the weight times the height that he lifts it.

So say he really does squat 300kg about 1m. The work done on the weight by the lifter is

W = F*d
W = mg*d
W = (300kg)(10m/s^2)(1m)
W = 3000 J

 

[minutemen11]

Yep, a muscle man that squats 300kg is doing zero net work. Tell them that next time you're at the gym

This makes no sense, PE+ KE= work. not just KE. PE work is mechanical work, there is 0 NET WORK done after the lifter puts down the weight because the net change in height is now 0.

I still feel that "work" is done opposing the weight of a dumbbell when it is held above you.. i need a mathematical expression...

 

[G1SG2]

This makes no sense, PE+ KE= work. not just KE. PE work is mechanical work, there is 0 NET WORK done after the lifter puts down the weight because the net change in height is now 0.

I still feel that "work" is done opposing the weight of a dumbbell when it is held above you.. i need a mathematical expression...

There is no work done when you're holding something above you. Work is done by a force when it causes a displacement. W=F*D. There is no displacement here. Therefore, the work done is zero.

There are forces involved, such as the force of gravity pushing downwards and the normal force between your hands and the weight, but there is no displacement from either side. No displacement=no work at all.

"there is 0 NET WORK done after the lifter puts down the weight because the net change in height is now 0"

No, the net change in height is not 0; it is the distance from the top where the weight was held all the way to the to point where the weight was put down. That's your delta h.

 

[minutemen11]

ok thanks

 

[sv3]

It can have potential energy as well.

If I'm holding a box steady at 1m and raise it to 2m, the net work done on the object is zero because it experienced zero change in it's kinetic energy even though it gained potential energy.

Can you explain how there is no net work done here?
You've applied a force over a distance. There is now greater PE than there was at 1m. If there is greater PE, work was put in to get it.

So how come there's no net work?

 

[sv3]

There is no work done when you're holding something above you. Work is done by a force when it causes a displacement. W=F*D. There is no displacement here. Therefore, the work done is zero.

There are forces involved, such as the force of gravity pushing downwards and the normal force between your hands and the weight, but there is no displacement from either side. No displacement=no work at all.

"there is 0 NET WORK done after the lifter puts down the weight because the net change in height is now 0"

No, the net change in height is not 0; it is the distance from the top where the weight was held all the way to the to point where the weight was put down. That's your delta h.

Just to clarify: If a weightlifter grabs a weight off the floor, raises it up over his head, and lowers it again, no work is done. PE is a conservative force and path-independent.

Where is this wrong?

 

[Seraph 84]

This makes no sense, PE+ KE= work. not just KE. PE work is mechanical work, there is 0 NET WORK done after the lifter puts down the weight because the net change in height is now 0.

I still feel that "work" is done opposing the weight of a dumbbell when it is held above you.. i need a mathematical expression...

Trust me, there's zero net work when he lifts the weight. The work done by the lifter will be 3000J, but the work done by gravity is -3000J because the force is opposite the lifter's force. The net work is thus, zero. So, if you're just talking to a lay person, then the lifter does do work on the object, but the net work done to raise the weight 1m was zero because gravity was working against him.

If you want a real mathematical expression for work, here it is.

W = F dot d

The dot in the expression is the dot product between 2 vectors. That same expression can also be written as

W = F*d*cos(theta), where theta is the angle between the force and the displacement vector. If he's just holding the weight there above his head, then he does no work because the displacement vector is zero. Similarly, if he grabs, say, a 30kg dumbbell, and carries it about 5m, he still does no work on the dumbbell because the angle between the force and the displacement vector is ninety degrees. The cosine of 90 degrees is zero, so work done is still zero even in this case.

I know it seems silly to think of work like this but that's the way it is in a pure physical sense. Now he is definitely expending energy to contract his muscles to lift the weights and what not, but we're talking about the net work done on the object, not just the work done by the lifter; gravity does work too.

 

[sv3]

Trust me, there's zero net work when he lifts the weight. The work done by the lifter will be 3000J, but the work done by gravity is -3000J because the force is opposite the lifter's force. The net work is thus, zero. So, if you're just talking to a lay person, then the lifter does do work on the object, but the net work done to raise the weight 1m was zero because gravity was working against him.

If you want a real mathematical expression for work, here it is.

W = F dot d

The dot in the expression is the dot product between 2 vectors. That same expression can also be written as

W = F*d*cos(theta), where theta is the angle between the force and the displacement vector. If he's just holding the weight there above his head, then he does no work because the displacement vector is zero. Similarly, if he grabs, say, a 30kg dumbbell, and carries it about 5m, he still does no work on the dumbbell because the angle between the force and the displacement vector is ninety degrees. The cosine of 90 degrees is zero, so work done is still zero even in this case.

I know it seems silly to think of work like this but that's the way it is in a pure physical sense. Now he is definitely expending energy to contract his muscles to lift the weights and what not, but we're talking about the net work done on the object, not just the work done by the lifter; gravity does work too.

If your going to add in gravity then obviously its 0!! You need to define your system better. Most questions I've seen ask "what is the work done by the lifter" or "what is the work done by gravity". I don't see questions asking what's the work done by the entire universe. I think that's what you're doing here isn't it? I could be wrong but that's where I think the differences lie in this thread.

 

[Seraph 84]

Yep. This topic exploded into something way more complicated than it needed to be. The original question was about the work energy theorem and if it could be applied to a box sliding across a table against friction.

You can use the work energy theorem to find the net work on any object if you know its initial and final kinetic energies, regardless of its potential energy.

 

[sv3]

Yep. This topic exploded into something way more complicated than it needed to be. The original question was about the work energy theorem and if it could be applied to a box sliding across a table against friction.

You can use the work energy theorem to find the net work on any object if you know its initial and final kinetic energies, regardless of its potential energy.

see that's where I can't agree. Work energy theorm only uses Ke in its formula. If there's some energy lost to heat you're disregarding it. If you have PE as well, you're disregarding it - isn't this where TME comes in?

Can you give me an example where there's KE and PE and you only need to consider KE to get work done? IT would be helpful. My physics is rusty and I thought work energy theorm only applies when only kinetic energy is the only energy present

 

[Seraph 84]

see that's where I can't agree. Work energy theorm only uses Ke in its formula. If you have PE as well, you're disregarding it. This will give you an incorrect answer. That's where TME comes in.

The work energy theorem is derived from the simple force times displacement equation. The acceleration is substituted with a kinematic equation. Potential energy does not matter at all when using the work energy theorem.

I think you're getting hung up on the total mechanical energy changing when I lift a block 1m from the ground. Sure, the potential energy increases, but that doesn't mean that there was net work done on that block; the net work will always depend on the change in kinetic energy because it should intuitively take a net amount of work to accelerate an object.

Say I fire a bullet in the air with an initial velocity of 1500m/s. As it reaches its highest point, it will have a velocity of zero. The only force acting on the bullet during it's time to reach maximum height was gravity. In this case, the net work done on the bullet was only done by gravity because it was the only force acting upon it. The work energy theorem still holds. The bullet gained a potential energy equal to its initial kinetic energy, and the work done by gravity will be the negative of the change in potential energy.

You can use that equation for any energy problem that calls for it, I promise.

In response to your question:

Say I have a baseball in my hand about 1m off the ground and I let it rip about 2m off the ground. The net work that was performed on the baseball is, again, equal to its change in kinetic energy, or the force I applied over the distance to accelerate it. The potential energy at those moments won't affect the net work done on the baseball because the work I perform to raise the ball 1m will exactly cancel out the work performed by gravity.

 

[sv3]

The work energy theorem is derived from the simple force times displacement equation. The acceleration is substituted with a kinematic equation. Potential energy does not matter at all when using the work energy theorem.

I think you're getting hung up on the total mechanical energy changing when I lift a block 1m from the ground. Sure, the potential energy increases, but that doesn't mean that there was net work done on that block; the net work will always depend on the change in kinetic energy because it should intuitively take a net amount of work to accelerate an object.

Say I fire a bullet in the air with an initial velocity of 1500m/s. As it reaches its highest point, it will have a velocity of zero. The only force acting on the bullet during it's time to reach maximum height was gravity. In this case, the net work done on the bullet was only done by gravity because it was the only force acting upon it. The work energy theorem still holds. The bullet gained a potential energy equal to its initial kinetic energy, and the work done by gravity will be the negative of the change in potential energy.

You can use that equation for any energy problem that calls for it, I promise.

In response to your question:

Say I have a baseball in my hand about 1m off the ground and I let it rip about 2m off the ground. The net work that was performed on the baseball is, again, equal to its change in kinetic energy, or the force I applied over the distance to accelerate it. The potential energy at those moments won't affect the net work done on the baseball because the work I perform to raise the ball 1m will exactly cancel out the work performed by gravity.

I think I'm getting there.....slowly......god I've fallen so fast after starting bio (its sucking the life out of me). I think I get the work thing now as far as Ke is concerned.

But let's say you lift a block from one shelf to another, 1 m apart. You are doing work aren't you? (sure if you include gravity it all cancels but you yourself had to put in work right). That's why i bolded the last sentance above. I think I am almost there but I guess what's tripping me up is that you consider the work done by gravity to cancel out your work in raising the ball - and I think the opposing force of gravity made you do work (when something gets lifted, isn't the work by gravity negative?). Shoot.

I was going to use the example of a frisbee in the air, moving at some speed above the ground. So if you want to bring this frisbee down, the work it takes is only equal to its kinetic energy? No need to worry about its PE? I think that's what your saying right? I think i get that now.

sorry for the confusion...........but thanks for the explanation

 

[G1SG2]

I think I'm getting there.....slowly......god I've fallen so fast after starting bio (its sucking the life out of me). I think I get the work thing now as far as Ke is concerned.

But let's say you lift a block from one shelf to another, 1 m apart. You are doing work aren't you? (sure if you include gravity it all cancels but you yourself had to put in work right). That's why i bolded the last sentance above. I think I am almost there but I guess what's tripping me up is that you consider the work done by gravity to cancel out your work in raising the ball - and I think the opposing force of gravity made you do work (when something gets lifted, isn't the work by gravity negative?). Shoot.

I was going to use the example of a frisbee in the air, moving at some speed above the ground. So if you want to bring this frisbee down, the work it takes is only equal to its kinetic energy? No need to worry about its PE? I think that's what your saying right? I think i get that now.

sorry for the confusion...........but thanks for the explanation

Yeah, just think of a hot air balloon floating in air. It's not accelerating up or down. It's just floating, meaning the equal and opposite forces are BALANCED-force of gravity pointing down, and the buoyant force is pointing up. The NET force is zero, but there are still forces involved (buoyant force and gravitational force). Similarly, when you raise a ball, the NET work is zero, but there is still work involved (the work done by gravity, and the work you do by lifting).

 

[sv3]

Yeah, just think of a hot air balloon floating in air. It's not accelerating up or down. It's just floating, meaning the equal and opposite forces are BALANCED-force of gravity pointing down, and the buoyant force is pointing up. The NET force is zero, but there are still forces involved (buoyant force and gravitational force). Similarly, when you raise a ball, the NET work is zero, but there is still work involved (the work done by gravity, and the work you do by lifting).

Right thanks. I think my issue is i am always thinking about the work I do, not thinking about gravity. Usually it works fine as questions pertain to the person, or gravity. But in these cases i guess you consider both. I actually don't think i paid attention to this cancelling of forces when doing problems and i was just fine and dandy. A little bit of extra thinking leads to a whole lot of trouble sometimes....crap

 

[G1SG2]

Right thanks. I think my issue is i am always thinking about the work I do, not thinking about gravity. Usually it works fine as questions pertain to the person, or gravity. But in these cases i guess you consider both. I actually don't think i paid attention to this cancelling of forces when doing problems and i was just fine and dandy. A little bit of extra thinking leads to a whole lot of trouble sometimes....crap

I think you'll be fine, as most questions are not concerned about the net work since it's usually zero. You'll probably be asked about the work done by something on either direction (gravity, etc).

 

[G1SG2]

I think you can also use W=delta KE + deltaPE if you have potential energy AND kinetic energy. Of course, if you start and end at rest deltaKE=0, which would explain why the work is the same for moving a box up straight up on a platform or using an incline to do the job.

 

[sv3]

I think you can also use W=delta KE + deltaPE if you have potential energy AND kinetic energy. Of course, if you start and end at rest deltaKE=0, which would explain why the work is the same for moving a box up straight up on a platform or using an incline to do the job.

Well that's what I was saying earlier but Seraph 84 was convincing in that you don't need to consider PE - you can figure out Work by just using Ke. It's all in the recent posts.

Oh..wait...now are you talking about the work done by just a person?
See.....if I lift a box up, and place it on a shelf that is higher, it starts and ends at rest. But the work I did is equal to mgh right?

Now let's say i a launch a ball into the air. Let's say its 100 metres high after some time and still moving (I'm on steroids). Would I need to use PE and KE to get the work I did? Or does using Ke alone take care of that? This is the question that has been murdering me

 

[minutemen11]

ok guys, KE+PE= mechanical work, there is no way around this formula. PE HAS TO BE considered for the scope of MCAT problems. If the object is on the ground then obviously there is no PE work, if the object is moved up or down there IS WORK being done BY the person ON the object! For example a typical person climbing stairs does approx 3000 J OF WORK- i read this somewhere i forget where. THIS IS HOW YOU BURN CALORIES CLIMBING.

When your considering NET work, its a different story because you have to look at input vs output. a kid climbing the ladder to the slide has done PE work. when this kid slides down the slide, all the PE work (in the scope of MCAT) is converted to KE work therefore the NET work is 0; correct?

 

[G1SG2]

ok guys, KE+PE= mechanical work, there is no way around this formula. PE HAS TO BE considered for the scope of MCAT problems. If the object is on the ground then obviously there is no PE work, if the object is moved up or down there IS WORK being done BY the person ON the object! For example a typical person climbing stairs does approx 3000 J OF WORK- i read this somewhere i forget where. THIS IS HOW YOU BURN CALORIES CLIMBING.

When your considering NET work, its a different story because you have to look at input vs output. a kid climbing the ladder to the slide has done PE work. when this kid slides down the slide, all the PE work (in the scope of MCAT) is converted to KE work therefore the NET work is 0; correct?

I think the net work in that situation would be just the work done by gravity, which could be found via PE=KE.

 

[sv3]

now do me!
its three posts up with a beat-down

 

[G1SG2]

Well that's what I was saying earlier but Seraph 84 was convincing in that you don't need to consider PE - you can figure out Work by just using Ke. It's all in the recent posts.

Oh..wait...now are you talking about the work done by just a person?
See.....if I lift a box up, and place it on a shelf that is higher, it starts and ends at rest. But the work I did is equal to mgh right?

Now let's say i a launch a ball into the air. Let's say its 100 metres high after some time and still moving (I'm on steroids). Would I need to use PE and KE to get the work I did? Or does using Ke alone take care of that? This is the question that has been murdering me

Correct, w=deltaPE=mgh would give you the work you did on the box.

For the ball-it starts off with KE and 0 potential energy, at the top of the trajectory it would have maximum PE, and at the bottom it would have KE, so I think KEf=KEi in the absence of friction. If you know delta PE, you automatically know delta KE because PE is converted to KE.

But this is a projectile. We know that if you drop a ball vertically, it starts with PE mgh and gets converted into KE, and the velocity would be sqrt2gh. However, if you THREW the ball down and gave it an initial vertical velocity, you have to use w=delta PE + delta KE because the ball started off with initial PE and initial KE.

I think this is correct.

 

[sv3]

Correct, w=deltaPE=mgh would give you the work you did on the box.

For the ball-it starts off with KE and 0 potential energy, at the top of the trajectory it would have maximum PE, and at the bottom it would have KE, so I think KEf=KEi in the absence of friction. If you know delta PE, you automatically know delta KE because PE is converted to KE.

But this is a projectile. We know that if you drop a ball vertically, it starts with PE mgh and gets converted into KE, and the velocity would be sqrt2gh. However, if you THREW the ball down and gave it an initial vertical velocity, you have to use w=delta PE + delta KE because the ball started off with initial PE and initial KE.

I think this is correct.

Right. Thank you! It has the KE from the throw, plus more KE via the PE that it loses on the way down.