Calculus II. Force and Work

Calculus II. Force and Work

February 9, 2015 1 / 8

What is force?

Physics provides a very precise definition of force in the form of Newton’sSecond Law of Motion.

The total force acting on an object is equal to the product of its mass andits acceleration.

(Force) = (Mass)·(Acceleration)

Thus, if mass is measured in kilograms (kg) and acceleration is measuredin meters per second-squared (m/s2) then Force is measured inkg ·m

s2.

Kilogram meters per second squared. This unit is called a newton1N = 1kg ·m

s2.

One Newton of force is the force needed to accelerate a one kg object onemeter per second squared.ExampleAn 2 kg object starts at rest. A force of 1N acts on it from the left(pushing in the positive x-direction) After 3 seconds what is the velocity ofthe object? How far has it moved?

February 9, 2015 2 / 8

What is force?

Physics provides a very precise definition of force in the form of Newton’sSecond Law of Motion.The total force acting on an object is equal to the product of its mass andits acceleration.



s2.


s2.


February 9, 2015 2 / 8

What is force?



Thus, if mass is measured in kilograms (kg) and acceleration is measuredin meters per second-squared (m/s2) then Force is measured in

kg ·ms2

.Kilogram meters per second squared. This unit is called a newton1N = 1kg ·m

s2.


February 9, 2015 2 / 8

What is force?




s2.

Kilogram meters per second squared.

This unit is called a newton1N = 1kg ·m

s2.


February 9, 2015 2 / 8

What is force?




s2.


s2.


February 9, 2015 2 / 8

What is force?




s2.


s2.

One Newton of force is the force needed to accelerate a one kg object onemeter per second squared.

ExampleAn 2 kg object starts at rest. A force of 1N acts on it from the left(pushing in the positive x-direction) After 3 seconds what is the velocity ofthe object? How far has it moved?

February 9, 2015 2 / 8

What is force?




s2.


s2.


February 9, 2015 2 / 8

What is work?

Colloquially work is the amount of effort put into something.

Physics again gives a more precise definition. Work is the product of aforce and the distance over which it is applied.

The SI unit for work is a Joule: 1J = 1N ·m = 1kg ·m2

s2One Joule is the amount of work done when you apply a force of 1Newton over a distance of one meter.For example: The force applied by gravity on an object of mass M isgiven by

(−9.8ms2

)·M.

Take a 10 kilogram bowling ball. Drop it from 15 meters up.How much force does gravity exert on the ball?How much work has gravity done by the time the ball hits the ground?

February 9, 2015 3 / 8

What is work?

Colloquially work is the amount of effort put into something.Physics again gives a more precise definition. Work is the product of aforce and the distance over which it is applied.



(−9.8ms2

)·M.


February 9, 2015 3 / 8

What is work?


The SI unit for work is a Joule: 1J =

1N ·m = 1kg ·m2


(−9.8ms2

)·M.


February 9, 2015 3 / 8

What is work?


The SI unit for work is a Joule: 1J = 1N ·m =

1kg ·m2


(−9.8ms2

)·M.


February 9, 2015 3 / 8

What is work?



s2

One Joule is the amount of work done when you apply a force of 1Newton over a distance of one meter.For example: The force applied by gravity on an object of mass M isgiven by

(−9.8ms2

)·M.


February 9, 2015 3 / 8

What is work?



s2One Joule is the amount of work done when you apply a force of 1Newton over a distance of one meter.

For example: The force applied by gravity on an object of mass M isgiven by

(−9.8ms2

)·M.


February 9, 2015 3 / 8

What is work?




(−9.8ms2

)·M.


February 9, 2015 3 / 8

What about when force is not constant?

Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?

Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.So the total work ∆W=(force)·(distance) done on this interval is pretty close to 1 ·∆x .We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x . If we estimate thework over each of these, we will have an estimate of the total work! Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8


Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?

Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.So the total work ∆W=(force)·(distance) done on this interval is pretty close to 1 ·∆x .We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x . If we estimate thework over each of these, we will have an estimate of the total work! Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8


Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.

So the total work ∆W=(force)·(distance) done on this interval is pretty close to 1 ·∆x .We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x . If we estimate thework over each of these, we will have an estimate of the total work! Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8


Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.So the total work ∆W=(force)·(distance) done on this interval is pretty close to

1 ·∆x .We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x . If we estimate thework over each of these, we will have an estimate of the total work! Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8


Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.So the total work ∆W=(force)·(distance) done on this interval is pretty close to 1 ·∆x .

We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x . If we estimate thework over each of these, we will have an estimate of the total work! Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8


Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.So the total work ∆W=(force)·(distance) done on this interval is pretty close to 1 ·∆x .We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x .

If we estimate thework over each of these, we will have an estimate of the total work! Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8


Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.So the total work ∆W=(force)·(distance) done on this interval is pretty close to 1 ·∆x .We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x . If we estimate thework over each of these, we will have an estimate of the total work!

Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8


Suppose that a force is acting on an object, but that force is differentdepending on the position of the object. Can we compute work?Example Suppose that an object is moving from the point x = 0 to thepoint x = 2 (In Meters). Suppose that a force acts on that object, andthat F (x) = x2 (in Newtons) for all x . Can we compute the total amountof work done?Well, as position changes from x = 1 to x = 1 + ∆x Can we estimate howmuch work is done?The force F (x) = x is between 12 and (1 + ∆x)2 on this small interval.Since these are both so close to 11 = 1, take F = 1 on this small interval,and accept that we have a small error.So the total work ∆W=(force)·(distance) done on this interval is pretty close to 1 ·∆x .We can say the same thing at any starting point, though, not just x = 1.Break the whole interval into subintervals of length ∆x . If we estimate thework over each of these, we will have an estimate of the total work! Onthe next slide we will see a Riemann sum.

February 9, 2015 4 / 8

Work as an integralSuppose that an object is moving from the point x = 0 to the point x = 2(In Meters). Suppose that a force acts on that object, and that F (x) = x2

(in Newtons) for all x . Can we compute the total amount of work done?

Break the interval [0, 2] up into subintervals of length ∆x . If ∆x is smallenough, then we can estimate F (x) = x2 over the i ’th interval by takingF (xi ) for any xi in that interval.Over the i ’th interval the work = (Force)·(distance) done is close toF (xi ) ·∆x . Add these up to estimate the force done over the whole

interval: Work ∼∑i

F (xi ) ·∆x

Using a smaller ∆x will improve the estimate, so

Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?

The x2 is in Newtons. x is measured in meters:Work = 8

3 Newtons · meters = 83 Joules.

February 9, 2015 5 / 8


(in Newtons) for all x . Can we compute the total amount of work done?Break the interval [0, 2] up into subintervals of length ∆x . If ∆x is smallenough, then we can estimate F (x) = x2 over the i ’th interval by takingF (xi ) for any xi in that interval.

Over the i ’th interval the work = (Force)·(distance) done is close toF (xi ) ·∆x . Add these up to estimate the force done over the whole


F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8


(in Newtons) for all x . Can we compute the total amount of work done?Break the interval [0, 2] up into subintervals of length ∆x . If ∆x is smallenough, then we can estimate F (x) = x2 over the i ’th interval by takingF (xi ) for any xi in that interval.Over the i ’th interval the work = (Force)·(distance) done is close toF (xi ) ·∆x . Add these up to estimate the force done over the whole

interval:

Work ∼∑i

F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8



interval: Work ∼

∑i

F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8




F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8




F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8




F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8




F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8




F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8




F (xi ) ·∆x


Work = lim∑i

F (xi ) ·∆x =

∫ 2

0F (x)dx

In our case: Work =

∫ 2

0x2dx =

8

3. Units?



February 9, 2015 5 / 8

Work as an integral

Work =

∫ b

a(Force)dx

for you:Hang a spring from the ceiling, and dangle a weightfrom it.

This spring will naturally come to some restinglength.

According to Hooke’s law, if you stretch a spring a distance x meters fromits resting length then the spring pulls back with a force of k · x for someconstant k N

m .Question: Start with a spring at rest. Grasp the weight and pull it down atotal distance of d .How much work have you done?Hint: First, for any displacement x , what is F (x)?Integrate to find the total work. Your answer will involve d and k .

February 9, 2015 6 / 8

Work as an integral

Work =

∫ b

a(Force)dx

for you:Hang a spring from the ceiling, and dangle a weightfrom it.

This spring will naturally come to some restinglength.According to Hooke’s law, if you stretch a spring a distance x meters fromits resting length then the spring pulls back with a force of k · x for someconstant k N


February 9, 2015 6 / 8

Work as an integral

Work =

∫ b

a(Force)dx

for you:Hang a spring from the ceiling, and dangle a weightfrom it.This spring will naturally come to some restinglength.

According to Hooke’s law, if you stretch a spring a distance x meters fromits resting length then the spring pulls back with a force of k · x for someconstant k N


February 9, 2015 6 / 8

Work as an integral

Work =

∫ b

a(Force)dx

for you:Hang a spring from the ceiling, and dangle a weightfrom it.This spring will naturally come to some restinglength.According to Hooke’s law, if you stretch a spring a distance x meters fromits resting length then the spring pulls back with a force of k · x for someconstant k N

m .

Question: Start with a spring at rest. Grasp the weight and pull it down atotal distance of d .How much work have you done?Hint: First, for any displacement x , what is F (x)?Integrate to find the total work. Your answer will involve d and k .

February 9, 2015 6 / 8

Work as an integral

Work =

∫ b

a(Force)dx


m .

Question: Start with a spring at rest. Grasp the weight and pull it down atotal distance of d .How much work have you done?Hint: First, for any displacement x , what is F (x)?Integrate to find the total work. Your answer will involve d and k .

February 9, 2015 6 / 8

Work as an integral

Work =

∫ b

a(Force)dx


m .Question: Start with a spring at rest. Grasp the weight and pull it down atotal distance of d .How much work have you done?Hint:

First, for any displacement x , what is F (x)?Integrate to find the total work. Your answer will involve d and k .

February 9, 2015 6 / 8

Work as an integral

Work =

∫ b

a(Force)dx



February 9, 2015 6 / 8

A more involved problemSuppose you have a 50 meter rope hanging over the edge of a building.The density of the rope is 1 kg per meter. How much work do you need todo if you are to lift it so that it all lies on the top of the building?

First find force given any length of rope.If a length of ` is hanging over the edge of the building, then how muchmass is hanging over the edge? If you are to lift this, you must overcomethe force of gravity. How strong is the force of gravity on this rope? Whatis F (`)?

Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg

)·(density)·(length) =

9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm

)Lift the rope a length ∆`.If you lift the rope a distance of ∆`, then the small amount of work doneis:∆W = (Force)·(distance) ∼ (9.8 · ` ·

(Nm

)) · (∆`).

As ∆`→ 0 this gives an equality of differentials:dW = 9.8 · ` ·

(Nm

)d` On the next slide we integrate.

February 9, 2015 7 / 8

A more involved problemSuppose you have a 50 meter rope hanging over the edge of a building.The density of the rope is 1 kg per meter. How much work do you need todo if you are to lift it so that it all lies on the top of the building?First find force given any length of rope.If a length of ` is hanging over the edge of the building, then how muchmass is hanging over the edge? If you are to lift this, you must overcomethe force of gravity. How strong is the force of gravity on this rope? Whatis F (`)?

Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm


(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) =

9.8(

Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm


(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm


(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · `

= 9.8 · `(Nm


(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm

)

Lift the rope a length ∆`.If you lift the rope a distance of ∆`, then the small amount of work doneis:∆W = (Force)·(distance) ∼ (9.8 · ` ·

(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm

)Lift the rope a length ∆`.If you lift the rope a distance of ∆`, then the small amount of work doneis:

∆W = (Force)·(distance) ∼ (9.8 · ` ·(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm

)Lift the rope a length ∆`.If you lift the rope a distance of ∆`, then the small amount of work doneis:∆W = (Force)·(distance)

∼ (9.8 · ` ·(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm


(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm


(Nm

)) · (∆`).

As ∆`→ 0 this gives an equality of differentials:

dW = 9.8 · ` ·(Nm


February 9, 2015 7 / 8


Force = 9.8(

Nkg

)·(mass) = 9.8

(Nkg


9.8(

Nkg

)· 1kg

m · ` = 9.8 · `(Nm


(Nm

)) · (∆`).


(Nm


February 9, 2015 7 / 8

A more involved problem

Last slide: If W is the work done and ` is the length of rope hang in overthe edge of the building then we have an equality of differentials:dW = 9.8 · ` ·

(Nm

)d`

Integrate to get work done

The total work done is W =

∫ `=50

`=0dW =

∫ `=50m

`=09.8 · ` ·

(N

m

)d`

Doing the integration:

W = 9.8(Nm

) (`2

2

)`=50m

`=0= 9.8

(Nm

)· 2500m2

2 = 12250N m

Let’s do a quick check: Are the units correct? Work is measured in Joules.One Joule is one N m. Work done = 12250J

February 9, 2015 8 / 8



(Nm

)d`



∫ `=50

`=0dW =

∫ `=50m

`=09.8 · ` ·

(N

m

)d`


W = 9.8(Nm

) (`2

2

)`=50m

`=0= 9.8

(Nm

)· 2500m2

2 = 12250N m


February 9, 2015 8 / 8



(Nm

)d`



∫ `=50

`=0dW =

∫ `=50m

`=09.8 · ` ·

(N

m

)d`


W = 9.8(Nm

) (`2

2

)`=50m

`=0= 9.8

(Nm

)· 2500m2

2 = 12250N m


February 9, 2015 8 / 8



(Nm

)d`



∫ `=50

`=0dW =

∫ `=50m

`=09.8 · ` ·

(N

m

)d`


W = 9.8(Nm

) (`2

2

)`=50m

`=0=

9.8(Nm

)· 2500m2

2 = 12250N m


February 9, 2015 8 / 8



(Nm

)d`



∫ `=50

`=0dW =

∫ `=50m

`=09.8 · ` ·

(N

m

)d`


W = 9.8(Nm

) (`2

2

)`=50m

`=0= 9.8

(Nm

)· 2500m2

2 =

12250N m


February 9, 2015 8 / 8



(Nm

)d`



∫ `=50

`=0dW =

∫ `=50m

`=09.8 · ` ·

(N

m

)d`


W = 9.8(Nm

) (`2

2

)`=50m

`=0= 9.8

(Nm

)· 2500m2

2 = 12250N m

Let’s do a quick check: Are the units correct?

Work is measured in Joules.One Joule is one N m. Work done = 12250J

February 9, 2015 8 / 8



(Nm

)d`



∫ `=50

`=0dW =

∫ `=50m

`=09.8 · ` ·

(N

m

)d`


W = 9.8(Nm

) (`2

2

)`=50m

`=0= 9.8

(Nm

)· 2500m2

2 = 12250N m


February 9, 2015 8 / 8

Documents

Calculus II. Force and Work