# Calculate the work required to move the block

• AbbeyC172
In summary, the man exerts a force of 180 lbf on the crate and the coefficient of friction between the block and the plane is 0.2. The work required to move the block 100 ft is 2000 lbf in Btu.
AbbeyC172

## Homework Statement

A man weighing 180 lbf pushes a block weighing 100 lbf alone a horizontal plane. The dynamic coefficient of friction between the block and plane is 0.2. Assuming that the block is moving at a constant speed, calculate the work required to move the block a distance of 100 ft considering (a) the man and (b) the block as a system. Express your answers in both lbf x ft and Btu.

N=ma
m= w/g

## The Attempt at a Solution

I started out by trying to draw a picture. So I got:

F3(total box) = F1 + F2
F1= (m)(a) --> a= 0 since V is constant? So F1= 0
F2= 0.2 (100) = 20
F3= 0+20 = 20

W(total box) = (F)(d)= (20)(100) = 2000 lb(ft) or 2.57 Btu

That's the work for the box I think? But then how to add in the force of the man's weight? Thank you so much in advance.[/B]

What is the force exerted by the man on the crate? When the man is the system, what is the force exerted by the crate on the man?

kuruman said:
What is the force exerted by the man on the crate? When the man is the system, what is the force exerted by the crate on the man?

180 lbf is the force exerted on the crate by the man and the block pushes back on him. But how do I work that into work for the whole system? Thank you by the way :)

180 lbf is the weight of the man, i.e. the force that the Earth exerts on the man. In your "attempt at a solution" you show F2 = 20 and F3 = 20. I assume that these numbers are in lbf units, but can you explain what forces F1, F2 and F3 are?

kuruman said:
180 lbf is the weight of the man, i.e. the force that the Earth exerts on the man. In your "attempt at a solution" you show F2 = 20 and F3 = 20. I assume that these numbers are in lbf units, but can you explain what forces F1, F2 and F3 are?

F1 would be the man's force on the box. F2 I put as the downward force on the box. F3 is the total force on the box. It's just how I set up the drawing to try to understand it better but apparently did not work.

AbbeyC172 said:
F1 would be the man's force on the box. F2 I put as the downward force on the box. F3 is the total force on the box. It's just how I set up the drawing to try to understand it better but apparently did not work.
What you say doesn't make sense. If F1 is the man's force on the box and F1 = 0 as you show in your attempt, then the man is not pushing on the box. Also, you say F2 is a downward force. What kind of downward force? In this problem, the only downward forces are the weight of the man (180 lbf) and the weight of the crate (100 lbf). There is no downward 20 lbf force. I think you are confused, so let's backtrack.

Consider the horizontal direction. Can you list all the forces that act on the crate in the horizontal direction only?

kuruman said:
What you say doesn't make sense. If F1 is the man's force on the box and F1 = 0 as you show in your attempt, then the man is not pushing on the box. Also, you say F2 is a downward force. What kind of downward force? In this problem, the only downward forces are the weight of the man (180 lbf) and the weight of the crate (100 lbf). There is no downward 20 lbf force. I think you are confused, so let's backtrack.

Consider the horizontal direction. Can you list all the forces that act on the crate in the horizontal direction only?

There would be the force of the man pushing on the box and then the coefficient of friction?

Yes, the force of the man and the force of friction. How are the magnitudes and directions of these forces related and why?

kuruman said:
Yes, the force of the man and the force of friction. How are the magnitudes and directions of these forces related and why?

The force of the friction is pushing back on the force from the man. The total force has to always be bigger than the force of friction. To calculate the force on the block, could I use:
F=
(F)g ?

Because once I figure out the force on the block and the force on the man, couldn't I add them both together to get my total force? Then I would be able to calculate the work as the question requests?

Thank you for your patience with me.

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AbbeyC172 said:
The force of the friction is pushing back on the force from the man
Forces act on objects. Forces do not act on other forces.
AbbeyC172 said:
The total force has to always be bigger than the force of friction.
Why would you think this?
AbbeyC172 said:
To calculate the force on the block, could I use: ##F_f=\mu F_g##
[equation reformatted]
Yes, you can calculate the force of friction that way.

Edit: Though technically you would need to first reason that ##F_n=F_g##, that is that the normal force between block and surface is equal to the downward force of gravity on the block.
AbbeyC172 said:
Because once I figure out the force on the block and the force on the man, couldn't I add them both together to get my total force?
Forces that act on different objects do not add.

First you should know that you have access of a whole lot of mathematical symbols if you click on the Σ symbol, far right, under "Have something to add?"
1. In the equation F = μ (F)g, do the two F symbols stand for the same force or a different force?
2. The total force is better known as the "net force", i.e. the sum of all the forces. What do you think the net force is in this particular case?

AbbeyC172
jbriggs444 said:
Forces act on objects. Forces do not act on other forces.

Why would you think this?

Yes, you can calculate the force of friction that way.

Edit: Though technically you would need to first reason that ##F_n=F_g##, that is that the normal force between block and surface is equal to the downward force of gravity on the block.

Forces that act on different objects do not add.
I would think the friction would have to be smaller than the net force in order to move the box?

kuruman said:
First you should know that you have access of a whole lot of mathematical symbols if you click on the Σ symbol, far right, under "Have something to add?"
1. In the equation F = μ (F)g, do the two F symbols stand for the same force or a different force?
2. The total force is better known as the "net force", i.e. the sum of all the forces. What do you think the net force is in this particular case?

1. The first F would be the total force, so F total = μ (F, force of box?) (9.8 m/s^2) so Ft = 0.2 (100) (9.8m/s^2) ?
2. Would the net force be equal to the force of the man minus the coefficient of friction?

Thank you for the tip about the symbols. If you can't already tell, haha I'm quite new at this.

AbbeyC172 said:
I would think the friction would have to be smaller than the net force in order to move the box?
Net force, by definition, is what remains after all appropriate cancellation between applied forces, including friction. If the net force is nonzero, acceleration will occur: Fnet=ΣFapplied=ma.

AbbeyC172
AbbeyC172 said:
I would think the friction would have to be smaller than the net force in order to move the box?
AbbeyC172 said:
Assuming that the block is moving at a constant speed
That suggests that the net force is...

AbbeyC172 said:
1. The first F would be the total force, so F total = μ (F, force of box?) (9.8 m/s^2) so Ft = 0.2 (100) (9.8m/s^2) ?
No. The F on the left is the force of friction. The F on the right is the perpendicular force acting on the surface, also known as the normal force. I also recommend that you use different symbols for different quantities. More correctly, this equation should be Ffriction=μ N. To find this force you only N because μ is given to you. Note that the coefficient of friction μ is not a force in itself; it is part of the expression that allows you to find the force of friction. So what do you think is the force that acts perpendicular to the surface at the point where the crate touches the floor?

Also consider posts #14 and #15 by @haruspex and @jbriggs444 and then try to bring it all together.

kuruman said:
No. The F on the left is the force of friction. The F on the right is the perpendicular force acting on the surface, also known as the normal force. I also recommend that you use different symbols for different quantities. More correctly, this equation should be Ffriction=μ N. To find this force you only N because μ is given to you. Note that the coefficient of friction μ is not a force in itself; it is part of the expression that allows you to find the force of friction. So what do you think is the force that acts perpendicular to the surface at the point where the crate touches the floor?

Also consider posts #14 and #15 by @haruspex and @jbriggs444 and then try to bring it all together.

Normal force? 100 lbf because that is the weight of the crate and the floor exerts the same back on it? Wouldn't those forces cancel out? If using Fnet=ΣFapplied=ma, then could I convert the 180 lb man into 81.65 kg and multiply that by 9.8m/s^2?

jbriggs444 said:

That suggests that the net force is...

It would be zero? Now I am so confused.

AbbeyC172 said:
Normal force? 100 lbf because that is the weight of the crate and the floor exerts the same back on it?
That is correct. Because the normal force exerted by the floor on the crate and the force exerted by the Earth on the crate are equal and opposite, the sum of all the vertical forces is zero. This means that the acceleration in the vertical direction is zero. Therefore, the velocity does not change in the vertical direction which means that if it is zero to begin with it remains zero and the crate does not fly off the floor or sink into it.
AbbeyC172 said:
It would be zero? Now I am so confused.
It will be zero, yes. To unconfuse yourself think of what I just posted about the vertical direction because it also applies to the horizontal direction. If the acceleration is zero in the horizontal direction, this means that the velocity does not change in that direction. So if the man and crate are moving at 2 ft/s they will keep on moving at 2 ft/s; if they are moving at 0.5 ft/s they will keep on moving at 0.5 ft/s. Newton's First Law of motion says that an object will maintain its velocity if there is no unbalanced force acting on it. This does not mean that if the forces acting on an object are balanced (i.e. their sum is zero) the object must have zero velocity. It means that the object must have constant velocity. Zero is only a special case of "constant" as are 2 ft/s and 0.5 ft/s.

kuruman said:
That is correct. Because the normal force exerted by the floor on the crate and the force exerted by the Earth on the crate are equal and opposite, the sum of all the vertical forces is zero. This means that the acceleration in the vertical direction is zero. Therefore, the velocity does not change in the vertical direction which means that if it is zero to begin with it remains zero and the crate does not fly off the floor or sink into it.

It will be zero, yes. To unconfuse yourself think of what I just posted about the vertical direction because it also applies to the horizontal direction. If the acceleration is zero in the horizontal direction, this means that the velocity does not change in that direction. So if the man and crate are moving at 2 ft/s they will keep on moving at 2 ft/s; if they are moving at 0.5 ft/s they will keep on moving at 0.5 ft/s. Newton's First Law of motion says that an object will maintain its velocity if there is no unbalanced force acting on it. This does not mean that if the forces acting on an object are balanced (i.e. their sum is zero) the object must have zero velocity. It means that the object must have constant velocity. Zero is only a special case of "constant" as are 2 ft/s and 0.5 ft/s.

This is tremendously helpful! Thank you. So since the velocity is constant, then the acceleration must be zero. But in regards to what the problem asks: am I on the right track to getting the force needed for the calculations?

F(n)= m x a = F(n) = (81.65 kg) x 9.8 m/s^2 = 800.17 N
Then to calculate the force of friction, plug N into the formula you stated above : Ffriction=μ N = Ffriction = 0.2(800.17) = 160.03

I know that work = force x displacement but I'm not quite sure how to my force for this formula. Do I subtract the force of friction from my normal force?

Remember that the work done by a constant force is given by W = F d cosθ, where
F = magnitude of the force
d = magnitude of the displacement
cosθ = cosine of the angle between the force vector and the displacement
It is given that d = 100 ft. We also have identified a number of forces. These are
In the horizontal direction
Fman on block, x
Fblock on man, x
Ffloor on block, x
Ffloor on man, x
In the vertical direction
FEarth on man, y
FEarth on block, y
Ffloor on block, y

If you can find numerical values for these forces and the appropriate cosines, you will be able to find the work that these forces do on the block or the man. Can you do it?

On edit: Added missing forces to the list.
Second edit: Use lbf units for forces because that's what's given and work values should be in lbf⋅ft.

Last edited:
kuruman said:
Remember that the work done by a constant force is given by W = F d cosθ, where
F = magnitude of the force
d = magnitude of the displacement
cosθ = cosine of the angle between the force vector and the displacement
It is given that d = 100 ft. We also have identified a number of forces. These are
In the horizontal direction
Fman on block, x
Fblock on man, x
Ffloor on block, x
Ffloor on man, x
In the vertical direction
FEarth on man, y
FEarth on block, y
Ffloor on block, y

If you can find numerical values for these forces and the appropriate cosines, you will be able to find the work that these forces do on the block or the man. Can you do it?

On edit: Added missing forces to the list.
Second edit: Use lbf units for forces because that's what's given and work values should be in lbf⋅ft.

Cosine should be zero since we have no angle.

Fman on block, x: 180 lbf
Fblock on man, x: 180 lbf
Ffloor on block, x: mg = 100 lbf(9.8) = 980 N
Ffloor on man, x: = mg = 180 lbf(9.8) = 1764 N
In the vertical direction
FEarth on man, y: 180 lbf
FEarth on block, y: 100 lbf
Ffloor on block, y: 100 lbf

I refuse to give up on this problem but where does the friction coefficient fit into the above variables?

AbbeyC172 said:
Cosine should be zero since we have no angle.
The cosine of zero is not zero. What is it? At what angle(s) is the cosine zero?
Let's try again with the forces. Remember that force is a vector. Use the standard convention that horizontal forces to the right are positive and to the left are negative; vertical forces up are positive and down are negative. More points to be made
A. The conversion factor (you can get that on the web) is 1 lbf = 4.4822 N. To get the mass in kg from the weight in lbf, you need to multiply by 4.4822 and divide by 9.8.
B. Why do you say that Fman on block, x = 180 lbf and Ffloor on block, x = 100 lbf? You forgot the reasoning behind this. Look at the crate. There are two items in the Universe that exert a horizontal force on it, man and floor. The block is moving at constant velocity and this means that the net force on it must be zero. What does this say about the relation between Fman on block, x and Ffloor on block, x?

AbbeyC172 said:
I refuse to give up on this problem but where does the friction coefficient fit into the above variables?
I am with you, don't give up. To see where the coefficient fits, consider this: friction is a force. Which of the 7 forces that I listed in post #21 can also be called friction? Hints: Is friction a horizontal or a vertical force? What object in the Universe exerts it, man, floor or Earth?

kuruman said:
The cosine of zero is not zero. What is it? At what angle(s) is the cosine zero?
Let's try again with the forces. Remember that force is a vector. Use the standard convention that horizontal forces to the right are positive and to the left are negative; vertical forces up are positive and down are negative. More points to be made
A. The conversion factor (you can get that on the web) is 1 lbf = 4.4822 N. To get the mass in kg from the weight in lbf, you need to multiply by 4.4822 and divide by 9.8.

B. Why do you say that Fman on block, x = 180 lbf and Ffloor on block, x = 100 lbf? You forgot the reasoning behind this. Look at the crate. There are two items in the Universe that exert a horizontal force on it, man and floor. The block is moving at constant velocity and this means that the net force on it must be zero. What does this say about the relation between Fman on block, x and Ffloor on block, x?I am with you, don't give up. To see where the coefficient fits, consider this: friction is a force. Which of the 7 forces that I listed in post #21 can also be called friction? Hints: Is friction a horizontal or a vertical force? What object in the Universe exerts it, man, floor or Earth?
Ahh, the cosine for 0 degrees which is what we have on a horizontal plane is 1. The cosine would be zero at 90 degrees.

The mass of the man: after converting.. m = f/a = 800.68/9.8= 81.7

I'm sorry, I think I'm getting more confused. I'm really lost as to what I'm even trying to figure out now. I'm going to have to ask someone face-to-face. I understand friction must be negative as it is opposing the force applied. Friction is a horizontal force. I'm assuming friction would be the floor on block, floor on man?

So I worked with a tutor and we figured out that
Fn= mg =100 lbs
F= Ffriction = 20 lbs
Force of friction = 20
Force of the man on the box: 20 lbs

Work on box by the man: 20 (100) (1) = 2000 ft x lbs
Total force = 0
Work being done on the system is zero because W = 0 (100) (1) = 0

So work done by just the man on the box is 2000 and since the work done total is zero, work done by the system would have to be -2000 lbs?

For some reason, I feel this isn't right?

Thank you so much for your help earlier but I needed someone present to talk it through to because this is such a weak subject for me to grasp,

AbbeyC172 said:
Friction is a horizontal force. I'm assuming friction would be the floor on block, floor on man?
Yes. Let's do first "floor on block". The relevant expression is Friction = Ffloor on block, x = μ Ffloor on block, y. Do you see how it works? So, what do you need to find the force of friction?
AbbeyC172 said:
I'm really lost as to what I'm even trying to figure out now.
Please answer my last question in B, in post #23. Then you will see better what's going on.

AbbeyC172 said:
friction would be the floor on block, floor on man?
You need to distinguish between kinetic and static friction. Which of those describes the friction between the floor and the man?
The surfaces in static friction do not move relative to each other, so static friction does no work.

AbbeyC172 said:
For some reason, I feel this isn't right?
Work done on a system is done by forces external to the system. The system cannot do work on itself. In this case if the system is block, then one has to consider the works done by all the forces on the block and add them up. That's what I was trying to guide you to. Likewise, if the system is the man, then one has to consider the works done by all the forces on the man and add them up.
AbbeyC172 said:
Work on box by the man: 20 (100) (1) = 2000 ft x lbs
Yes.
AbbeyC172 said:
So work done by just the man on the box is 2000 and since the work done total is zero, work done by the system would have to be -2000 lbs?
Yes, if you you rephrase it as the work done on the man.
AbbeyC172 said:
Thank you so much for your help earlier but I needed someone present to talk it through to because this is such a weak subject for me to grasp,
I hope you have better understanding of the subject now. Face to face tutoring can be very productive.

AbbeyC172
kuruman said:
Work done on a system is done by forces external to the system. The system cannot do work on itself. In this case if the system is block, then one has to consider the works done by all the forces on the block and add them up. That's what I was trying to guide you to. Likewise, if the system is the man, then one has to consider the works done by all the forces on the man and add them up.

Yes.

Yes, if you you rephrase it as the work done on the man.

I hope you have better understanding of the subject now. Face to face tutoring can be very productive.
Thank you for all the help! I will let you know what my professor says when he grades the paper.

## 1. What is the formula for calculating work?

The formula for calculating work is W = F x d, where W represents work, F represents the applied force, and d represents the distance over which the force is applied.

## 2. How do I determine the force required to move the block?

The force required to move the block depends on the mass of the block and the acceleration of the block. You can calculate the force by multiplying the mass by the acceleration (F = m x a).

## 3. Can I use any unit of measurement for work?

No, work is typically measured in joules (J) in the metric system. In the imperial system, work is measured in foot-pounds (ft-lb) or pound-feet (lb-ft).

## 4. What factors can affect the amount of work required to move a block?

The amount of work required to move a block can be affected by the mass of the block, the distance it needs to be moved, and the presence of friction or other resistive forces.

## 5. Does the direction of the force affect the amount of work required?

Yes, the direction of the force does affect the amount of work required. Work is a scalar quantity, meaning it only takes into account the magnitude of the force and the distance, not the direction. However, the direction of the force can affect the amount of force needed to move the block and therefore impact the work required.

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