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 = Ax 2 + Ay 2 + Az 2 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
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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