Work and Kinetic Energy I

■ Scalar Product

The scalar product of any two vectors A and B is a scalar quantity equal to the product of the

magnitudes of the two vectors and the cosine of the angle φ between them

A◦B ≡ ABcosφ

The unit vectors i, j and k lie in the positive x, y and z directions rectangle coordinate system.

Therefore,

i◦i = j◦j = k◦k = 1 i◦j = i◦k = j◦k = 0

If vector A and B is defined as A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk, then

A◦B = Ax Bx + Ay By + Az Bz

and A◦A = Ax2 + Ay

2 + Az2

Ex7.2) A = 2i + 3j and B = -i + 2j. (a) Find the scalar product of A◦B. (b) Find the angle φ

between A and B.

■ Work done by Constant Force

The work W done on an object by an agent exerting a constant force on the object is the product

of the component of the force in the direction of the displacement and the magnitude of the

displacement.

W = F ◦ d = Fdcosφ

- Unit of work is expressed in Joules(J)

- If you apply 1N of force over a distance of 1meter, then you are doing 1Joule of work.

FN

Fapp

φ

Fapp cos φ

d= distance traveled

Fg = mg

W = Fapp ◦ d = Fdcosφ

- Work done by normal force FN is zero; W=FN dcos90º =0

- Work done by gravitational force Fg (=weight) is zero; W=Fg dcos90º = 0

1. A boy exerts a horizontal force of 10N on a box at a distance of 25m. How much work did he

do on the box?

a) 200J b) 250J c) 300J d) 350J

2. A girl exerts a force of 35N at an angle of 35° above the horizontal along the handle of her

wagon as she pulls it 25m across a level sidewalk. How much work does she do on her wagon?

a) 717J b) 156J c) 134J d) 120J

3. An object of mass 3kg is dropped from a height of 4m above the ground. Find the work done

by gravity(Wg) throughout the motion until it reaches the ground.

a) 117.6J b) -117.6J c) 29.4J d) -29.4J

4. A box on the floor is lifted by man. If the mass of the box is 60kg, and he

lifts it 2m from the ground at a constant speed,

i) how much work did the man do on the box?

a) 588J b) -588J c) 1176J d) -1176J

FL

Fg

ii) how much work did gravity do on the box?

a) 588J b) -588J c) 1176J d) -1176J

5. A crate of mass 5kg is released from the top of

frictionless ramp. The angle of the incline is 18°.

i) What is the force that does work on the crate?

ii) Find the work done on the crate over distance of 4m along the ramp.

a) 34.4J b) 47.3J c) 51.2J d) 60.6J

■ Work-Kinetic Energy Theorem

If you apply a force over a certain distance, there will be a change in kinetic energy

vi distance ‘d’ vf

Fapp Fapp

Ki + F d cosφ = Kf

1

2mvi

2 +Fapp dcosφ = 1

2mvf

2

Ki + 𝜮W = Kf : Work-Kinetic Energy Theorem where K= 1

2mv2

6. A 150g mass has a velocity v=(2i + 6j) m/s at a certain instant. What is its kinetic energy?

( Use vector dot product to find the speed. That is, v◦v=v2)

7. How much horizontal force is needed to accelerate a 1300kg car from rest to a speed of 20m/s

in a distance of 80m?

8. Revisit question 4 using Work-Kinetic Energy Theorem

A box on the floor is lifted by man. If the mass of the box is 60kg, and he lifts

it 2m from the ground at a constant speed,

i) how much work did the man do on the box?

a) 588J b) -588J c) 1176J d) -1176J

FL

Fg

ii) how much work did gravity do on the box?

a) 588J b) -588J c) 1176J d) -1176J

W= F ◦ d f =µFN K =

1

2mv2 Ki + 𝛴W = Kf

9. An object of mass 2kg is moving on a frictionless surface with a constant speed of 8m/s. How

much force is needed to stop the object over a distance of 10m?

a) 3.9N b) 6.4N c) 8.3N d) 11.8N

10. A 4kg box is sliding on a surface with an initial speed of 8m/s. The coefficient of kinetic

friction between the box and the floor is µk =0.4. What is the speed of the box after it moved a

distanced of 4m? (*~frictional force f=µFN, where FN=mg in this case ~*)

a) 19.6m/s b) 14.2m/s c) 10.5m/s d) 5.7m/s

11. A 8kg box is released from rest on a 30° inclined frictionless plane and accelerates down the

incline. If the distance from the block to the end of the ramp is 4m, what is the speed of the block

when it reaches the bottom of the ramp?

a) 6.3m/s b) 3.6m/s c) 1.6m/s d) 0.9m/s

W= F ◦ d f =µFN K =

1

2mv2 Ki + 𝛴W = Kf

12. A 8kg box is released from rest on a 30° inclined rough plane and accelerates down the

incline. The coefficient of friction µk=0.54.

i) If the distance from the block to the end of the ramp is 4m, what is the speed of the block when

it reaches the bottom of the ramp?

a) 3.3m/s b) 2.4m/s c) 1.6m/s c) 0.4m/s

ii) The block will eventually slide off the ramp and continue on the surface. If the coefficient of

kinetic friction on the surface was µk=0.12, how far will the block travel until it stops?

a) 2.21m b) 1.63m c) 1.07m d) 0.64m

W= F ◦ d f =µFN K =

1

2mv2 Ki + 𝛴W = Kf

13. A 12kg crate is pushed up on a frictionless

incline plane from the bottom to top at constant

speed. The height of the incline is h=5m and

the distance of the plane is d=14m. The angle

of the incline is 21°. How much work was done

by the pushing force Fp? That is, find Wp.

a) 590J b) 1200J c) -590J d) -1200J

d=14m

Fp

h=5m

Hint:

Fp

φ

θ

Fg

- Use Ki + 𝛴W = Kf

- Work done by gravity is Wg = Fgdcosφ, where φ= θ+90°.

- Work done by pushing force Wp= Fpcos0°×d.

14. A crate of mass 10kg is pulled up a rough

incline plane with an initial speed of 1.5m/s. The

pulling force is 100N parallel to the incline, which

makes an angle of 20° with the horizontal. The

coefficient of kinetic friction is 0.4, and the crate

is pulled 5m.

(a) Find the work done by the pulling force(Wp)

FN

Fp

fk

Fg

(b) Find the work done by the gravitational force(Wg)

(c) Find the work done by the frictional force(Wf )

(d) Find the work done by the normal force(WN )

(e) Find the change in kinetic energy

**(f) Find the energy lost due to friction?

Solution for 14.

(a)Begin by writing

Ki + 𝛴W = Kf

Identify the forces acting on the object by drawing free-body diagram as above. All the forces has

the potential to the work on the crate, so

Ki + Wp + Wg + Wf + WN = Kf

Wp= Fp dcos0° = Fp d=100×5=500J

(b) Wg= Fg dcos(90°+20°) = mgdcos(110°)= –167.6J

(c) Wf = fk dcos(180°) = –μkmgdcosθ = –0.4×10×9.8×5×cos20°= –184.2J

(d) WN = FN dcos(90°) = 0

(e) From Ki + Wp + Wg + Wf = Kf ,

ΔK = Kf – Ki = Wp + Wg + Wf =500 –167.6 –184.2 =148.2J

(f) energy loss = work done by friction. So Eloss = Wf = –184.2J

■ Work done by Varying Force

The work done on a particle by a force that varies according to position F(x) is expressed as

W = ∫ 𝐹𝑥 𝑑𝑥𝑥𝑓

𝑥𝑖

If more than one force acting on the particle does work on it, then

∑W = ∫ ( ∑𝐹𝑥) 𝑑𝑥𝑥𝑓

𝑥𝑖

Example 7.4 Calculating Total Work Done from the Graph. see p.184 in the textbook

A force acting on a particle varies with x.

Calculate the work done by the force as the

particle moves from x=0 to x=6m.

Fx(N)

5

x(m)

1 2 3 4 5 6

Sol)

Method I ) The work done by the force is equal to the area under the curve from x=0 to x=6m.

The area from x=0 to x=4m is 5×4 = 20J and the area from x=4m to x=6m is ½×2×5=5J

So

Wtot = 20 + 5 =25J

Method II ) The force acting on the particle from x=0 to x=4m is constant.

So

W = Fd = 5×4 = 20J

However, the force acting on the particle from x=4m to x=6m decreases as a function of position.

The linear function only from x=4m to 6m can be expressed as

F(x) = −5

2𝑥 + 15

So W = ∫ 𝐹𝑥 𝑑𝑥𝑥𝑓

𝑥𝑖=∫ (−

5

2𝑥 + 15) 𝑑𝑥

6

4 =( −

5

4𝑥2 + 15𝑥) ]

64

=5J

So the total work done on the particle is

Wtot = 20 + 5 =25J

W= F ◦ d f =µFN K =

1

2mv2 Ki + 𝛴W = Kf

15. The force acting on a particle is F(x) = (8x–16)N, where x is in meters. Find the work done by

this force as the particle moves from x =0 to x =3m.

a) +12J b) –12J c) +24J d) –24J

16. A force F=(4xi + 3yj) N acts on an object as it moves in the direction from the origin to x=5m.

Find the work done on the object by the force.

a) +25J b) –25J c) +50J d) –50J

■ Work done by the Spring

A block on a horizontal, frictionless surface is connected to a spring. If the spring is either

stretched or compressed a small distance from its unstretched(=equilibrium) configuration, its

exerts on the block a force of magnitude

Fs= –kx

where x is the displacement of the block from its unstretched(x=0) position and k is a positive

constant called force constant of the spring. In other words, k represents the stiffness of the spring.

i) The work done by the spring on the block

Ws = ∫ 𝐹𝑠 𝑑𝑥𝑥𝑓

𝑥𝑖= ∫ (−𝑘𝑥) 𝑑𝑥

𝑥𝑓

𝑥𝑖 = −

1

2𝑘𝑥𝑓

2 − (−1

2𝑘𝑥𝑖

2) = 1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2

=> Ws = 𝟏

𝟐𝒌𝒙𝒊

𝟐 −𝟏

𝟐𝒌𝒙𝒇

𝟐

ii) The work done by the applied force on the block

A block is pulled from xi = 0 to xf = xmax on a frictionless surface

by a force Fapp. If the process is carried our very slowly, the

applied force is equal and opposite of the spring force at all

times. That is, at any position, the applied force Fapp is equal to

and opposite of the spring force Fs, so that

Fs Fapp

xi = 0 xf = xmax

Fapp = –(–kx) = kx. Therefore, the work done by the applied force is

Wapp = ∫ 𝐹𝑎𝑝𝑝 𝑑𝑥𝑥𝑚𝑎𝑥

0= ∫ 𝑘𝑥 𝑑𝑥

𝑥𝑚𝑎𝑥

0 =

1

2𝑘𝑥𝑚𝑎𝑥

2

---------------------------------------------------------------------------------------------------------------------

Ki + 𝛴W = Kf Ws = 1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2

Example 7.5 Measuring k for a Spring see p.187

A common technique used to measure the force constant of a spring is demonstrated by hanging

the spring vertically and attaching an object of mass m to its lower end. Under the action of the

“load” mg, the spring stretches a distance d from its equilibrium position.

(A) If the spring is stretched 2.0cm by a suspended object having a mass of 0.55kg, what is the

force constant of the spring?

(B) How much work is done by the spring on the object as it stretches through this distance?

W =∫ 𝑭 ◦ dr f =µFN K = 1

2mv2 Fs= –kx Ws =

1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 Ki + 𝛴W = Kf

17. A 10kg block on a horizontal frictionless surface is attached

to a light spring(force constant k=1200N/m). The block is

initially at rest at its equilibrium position when a force P acting

parallel to the surface is applied to the block. When the block is

8cm from the equilibrium position, it has a speed of 0.8m/s.

P

i) How much work was done by the force P as the block moves the 8cm?

a) 3.84J b) 5.54J c) 7.04J d) 12.52J

ii) How much work was done by the spring as the block moves the 8cm?

a) -3.84J b) -5.54J c) -7.04J d) -12.52J

18. A block of mass 2kg slides on a frictionless surface.

The speed of the block before it touches the spring is 6m/s.

How fast is the block moving at the instant the spring has

been compressed 15cm? k=2000N/m

k

a) 3.7m/s b) 4.4m/s c) 4.9m/s d) 5.4m/s

W =∫ 𝑭 ◦ dr f =µFN K = 1

2mv2 Fs= –kx Ws =

1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 Ki + 𝛴W = Kf

19. A block of mass 1.6kg is attached to a horizontal spring that

a force constant of 1×103N/m. The spring is compressed 2cm

and is then released from rest. See solution below

(i) Calculate the speed of the block as it passes through the

equilibrium position x =0 if the surface is frictionless.

2cm

x=0

a) 1.94m/s b) 1.38m/s c) 0.50m/s d) 0.39m/s

(ii) Calculate the speed of the block as it passes through the equilibrium position x =0 if a

constant frictional force of 4N retards the motion from the moment it is released.

a) 1.94m/s b) 1.38m/s c) 0.50m/s d) 0.39m/s

------------------------------------------------------------------------------------------------------------------

solution for 19.

(i) Use Ki + 𝛴W = Kf . The only force doing work on the block is the spring force, so

Ki + Ws = Kf

Since Ws = 1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2

1

2mvi

2 + 1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 = 1

2mvf

2

Our observation begins when the block is released, so xi=0.02m and since we are trying to find

the speed the instant the block passes x=0, we set xf =0. Since vi =0 and solve for vf , we get

vf =0.5m/s

(ii) There are two forces doing work on the block. The spring force and the frictional force, so

Ki + Ws + Wf = Kf

Since Ws = 1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 and Wf =fkcos180°d, where d=xi , we get

1

2mvi

2 + 1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 – fk xi= 1

2mvf

2

So vf =0.39m/s

----------------------------------------------------------------------------------------------------------------

W =∫ 𝑭 ◦ dr f =µFN K = 1

2mv2 Fs= –kx Ws =

1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 Ki + 𝛴W = Kf

20. A block of mass 2kg is attached to a horizontal spring that a force constant of 500N/m. The

spring is pulled 5cm to the right and is then released from rest.

i) Calculate the speed of the block as it passes through the equilibrium position x=0 if the surface

is frictionless.

a) 3.45m/s b) 1.84m/s c) 0.79m/s d) 0.53m/s

ii) Calculate the speed of the block as it passes through the equilibrium position x=0 if the

coefficient of kinetic friction between the block and the surface is 0.35.

a) 3.45m/s b) 1.84m/s c) 0.79m/s d) 0.53m/s

21. A block attached to a spring on a frictionless horizontal surface is being pulled at a constant

speed by a boy. If the spring constant is 575N/m and the block is pulled 0.4m from the

equilibrium position, then how much work was done on the block by the boy?

a) 40J b) 46J c) 55J d) 68J e) 73J