1. Kinetic Energy and Work
Chapter 7
Kinetic Energy and Work

2. Work and energy are scalars, measured in N·m or Joules, J, 1J=kg∙m2/s2
Energy: scalar quantity associated with a state (or condition) of one or more objects.
Work and energy are scalars, measured in N·m or Joules, J, 1J=kg∙m2/s2
Energy can exist in many forms - mechanical, electrical, nuclear, thermal, chemical….

3. Energy can be converted from one type to another but never destroyed.
Work and Energy
Energy is a conserved quantity - the total amount of energy in the universe is constant.
Energy can be converted from one type to another but never destroyed.
Work and energy concepts can simplify solutions of mechanical problems - they can be used in an alternative analysis

4. Work done by a constant force
Work done on an object by a constant force is defined to be the product of the magnitude of the displacement and the component of the force parallel to the displacement
Where FII is the component of the force F parallel to the displacement d
W = FII · d

5. Work 1J=kg∙m2/s2 In other words - Where q is the angle between F and d
If q is > 90o, work is negative. A decelerating car has negative work done on it by its engine.
The unit of work is called Joule (J), 1 J = 1 N·m
1J=kg∙m2/s2
W = F d cosq q d F

6. Work - on and by
A person pushes block 30 m along the ground by exerting force of 25 N on the trolley. How much work does the person do on the trolley?
W = F d = 25N x 30m = 750 Nm
Trolley does -750 Nm work on the person
Ftp
Fpt d

7. Mechanical Energy
Mechanical energy (energy associated with masses) can be thought of as having two components: kinetic and potential
Kinetic energy is energy of motion
Potential energy is energy of position

8. Kinetic Energy
In many situations, energy can be considered as “the ability to do work”
Energy can exist in different forms
Kinetic energy is the energy associated with the motion of an object
A moving object can do work on another object
E.g hammer does work on nail.

9. Kinetic Energy
Consider an object with mass m moving in a straight line with initial velocity vi. To accelerate it uniformly to a speed vf a constant net force F is exerted on it parallel to motion over a distance d.
Work done on object
W = F d = m a d (NII) So

10. Kinetic Energy
If we rearrange this we obtain
We define the quantity ½mv2 to be the translational kinetic energy (KE) of the object
This is the ‘Work-Energy Theorem’:
“The net work done on an object is equal to its change in kinetic energy”
W = DKE

11. The Work-Energy Theorem
W = DKE
Note
The net work done on an object is the work done by the net force.
Units of energy and work must be the same (J)

12. Energy: scalar quantity associated with a state (or condition) of one or more objects.
Kinetic energy.
Work.
Work - Kinetic energy theorem.
Work done by a constant force
- Gravitational force
V. Work done by a variable force.
- Spring force.
1D-Analysis
3D-Analysis
VI. Power

13. I. Kinetic energy II. Work Units:
Energy associated with the state of motion of an object.
Units:
1 Joule = 1J = 1 kgm2/s2
II. Work
To +W
From -W
Energy transferred “to” or “from” an object by means of a force acting on the object.

14. - Constant force:
Work done by the force = Energy transfer due to the force.

15. To calculate the work done on an object by a force during
a displacement, we use only the force component along
the object’s displacement. The force component perpendicular
to the displacement does zero work.
Assumptions: 1) F = constant force
2) Object is particle-like (rigid object, all parts
of the object must move together).
A force does +W when it has a vector component in the same direction as the displacement, and –W when it has
a vector component in the opposite direction. W=0 when it has no such vector component.

16. Net work done by several forces = Sum of works done by
individual forces.
Calculation: 1) Wnet= W1+W2+W3+…
2) Fnet  Wnet=Fnet d

17. II. Work-Kinetic Energy Theorem III. Work done by a constant force
Change in the kinetic energy of the
particle = Net work done on the particle
III. Work done by a constant force
- Gravitational force:
Rising object: W= mgd cos180º = -mgd 
Fg transfers mgd energy from the object’s kinetic energy.
Falling down object: W= mgd cos 0º = +mgd 
Fg transfers mgd energy to the object’s kinetic energy.

18. External applied force + Gravitational force:
Object stationary before and after the lift: Wa+Wg=0
The applied force transfers the same amount of energy to the object as the gravitational force transfers from the object.

19. IV. Work done by a variable force
- Spring force:
Hooke’s law
k = spring constant  measures spring’s stiffness. Units: N/m

20. Work done by a spring force:
Hooke’s law: Fx xi Δx xf x Fj
Work done by a spring force:
Spring is massless  mspring << mblock
Ideal spring  obeys Hooke’s law exactly.
Contact between the block and floor is frictionless.
Block is particle-like.
Assumptions:

21. The block displacement must be divided
- Calculation:
The block displacement must be divided
into many segments of infinitesimal width, Δx.
2) F(x) ≈ cte within each short Δx segment. Fx xi Δx xf x Fj
Ws>0  Block ends up closer to the relaxed position (x=0) than it was initially.
Ws<0  Block ends up further away from x=0.
Ws=0  Block ends up at x=0.

22. Work done by an applied force + spring force:
Block stationary before and after the displacement:
Wa= -Ws  The work done by the applied force displacing
the block is the negative of the work done by the spring force.

23. Work done by a general variable force:
1D-Analysis
Geometrically: Work is the area between the curve F(x)
and the x-axis.

24. Work-Kinetic Energy Theorem - Variable force
3D-Analysis
Work-Kinetic Energy Theorem - Variable force

25. V. Power Time rate at which the applied force does work.
- Average power: amount of work done
in an amount of time Δt by a force.
Instantaneous power:
instantaneous time rate of doing work.
Units: 1 watt= 1 W = 1J/s
1 kilowatt-hour = 1000W·h = 1000J/s x 3600s = 3.6 x 106 J
= 3.6 MJ F φ x

26. (a) downward angle θ as the block moves right-ward through 1m across
N2. In the figure (a) below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m across
a frictionless floor. Find an expression for the speed vf at the end of
that distance if the block’s initial velocity is:
(a) 0 and (b) 1m/s to the right. (c) The situation in Fig.(b) is similar in that
the block is initially moving at 1m/s to the right, but now the 2N force
is directed downward to the left. Find an expression for the speed of
the block at the end of the 1m distance.
(a) N mg Fx Fy N Fx Fy mg

27. downward angle θ as the block moves right-ward through 1m across
N2. In the figure (a) below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m across
a frictionless floor. Find an expression for the speed vf at the end of
that distance if the block’s initial velocity is:
(a) 0 and (b) 1m/s to the right. (c) The situation in fig.(b) is similar in that
the block is initially moving at 1m/s to the right, but now the 2N force
is directed downward to the left. Find an expression for the speed of
the block at the end of the 1m distance. N mg Fx Fy N Fx Fy mg

28. N6. A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface. Beginning
at t=0, a steady wind pushes on the lunchbox in the negative
direction of x, Fig. below. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force
from the wind do on the lunch box from t=1s to t=5s?

29. N6. A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface. Beginning
at t=0, a steady wind pushes on the lunchbox in the negative
direction of x, Fig. below. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force
from the wind do on the lunch box from t=1s to t=5s?

30. N6. A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface. Beginning
at t=0, a steady wind pushes on the lunchbox in the negative
direction of x, Fig. below. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force
from the wind do on the lunch box from t=1s to t=5s?

31. N12. In the figure below a horizontal force Fa of magnitude 20N is applied to a 3kg book, as the book slides a distance of d=0.5m up a frictionless ramp. (a) During the displacement, what is the net force done on the book by Fa, the gravitational force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement? x y mg N
Fgy
Fgx

32. N12. In the figure below a horizontal force Fa of magnitude 20N
is applied to a 3kg book, as the book slides a distance of d=0.5m
up a frictionless ramp. (a) During the
displacement, what is the net force done
on the book by Fa, the gravitational force
on the book and the normal force on the
book? (b) If the book has zero kinetic
energy at the start of the displacement,
what is the speed at the end of the
displacement? y x mg N
Fgy
Fgx

33. N12. In the figure below a horizontal force Fa of magnitude 20N
is applied to a 3kg book, as the book slides a distance of d=0.5m
up a frictionless ramp. (a) During the
displacement, what is the net force done
on the book by Fa, the gravitational force
on the book and the normal force on the
book? (b) If the book has zero kinetic
energy at the start of the displacement,
what is the speed at the end of the
displacement? y x mg N
Fgy
Fgx

34. N15. (a) Estimate the work done represented by the graph below
in displacing the particle from x=1 to x=3m. (b) The curve is given
by F=a/x2, with a=9Nm2. Calculate the work using integration

35. N19. An elevator has a mass of 4500kg and can carry a maximum
load of 1800kg. If the cab is moving upward at full load at constant
speed 3.8m/s, what power is required of the force moving the cab
to maintain that speed? Fa mg

36. N17. A single force acts on a body that moves along an x-axis.
The figure below shows the velocity component versus time for
the body. For each of the intervals
AB, BC, CD, and DE, give the sign
(plus or minus) of the work done by
the force, or state that the work is zero. v B C D A t E

37. 15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
work done on the canister by (c) your
force (via the cord) and (d) the
gravitational force on the canister? P2 T T T P1 mg

38. 15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
work done on the canister by (c) your
force (via the cord) and (d) the
gravitational force on the canister? P2 T T T P1 mg
(b) To rise “m” 0.02m, two segments of the cord must be shorten by that
amount. Thus, the amount of the string pulled down at the left end is: 0.04m

39. There is no change in kinetic energy.
15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
work done on the canister by (c) your
force (via the cord) and (d) the
gravitational force on the canister? P2 T T T P1 mg
WF+WFg=0
There is no change in kinetic energy.

40. From the point of view of kinetics, does the relative position of the
Challenging problems – Chapter 7
Two trolleys of masses m1=400 kg and m2=200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N.
From the point of view of kinetics,
does the relative position of the
trolleys matter? If the rod can
only stand an applied force of 500N,
which trolley should be up front? N1
Situation 1 N2
F1r
F2r F
m1g
m2g
Action and reaction forces: F1r force that the rod does on m1
F2r force that the rod does on m2
F’2r
F’1r
F’1r force that m1 does on rod.
F’2r force that m2 does on rod.
Situation 1
Rod negiglible mass, Fnet on rod=0
F’1r = F’2r  rigid rod does not deform
F1r = F’1r  Action / Reaction
Forces that the rod does on the blocks as it tries to counteract the deformation that F could induce.

41. Challenging problems – Chapter 7
Two trolleys of masses m1=400 kg and m2=200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N. From the point of view of kinetics, does the relative position of the trolleys matter? If the rod can only stand an applied force of 500N, which trolley should be up front?
F’2r
F’1r
Action and reaction forces: F1r force that the rod does on m1
F2r force that the rod does on m2
Situation 2
F’1r force that m1 does on rod.
F’2r force that m2 does on rod.
Rod negiglible mass, Fnet on rod=0
F’1r = F’2r  rigid rod does not deform
F1r = F’1r  Action / Reaction
Situation 2 N2 N1
From kinetic point of view, Situation1=Situation2same acceleration”.
F2r
F1r F
From dynamics  F1r(1)≠F1r(2) Only situation (1) is possible F1r=F’1r = 400N <Fmax=500N
m1g
m2g

42. 2. A car with a weight of 2500N working with a power of 130kW develops a velocity of 112 km/hour when traveling along a horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not negligible: What is the value of the frictional forces?
(a) What is the car’s maximum velocity on a 50 incline hill? (b) What is the power if the car is traveling on a 100 inclined hill at 36km/h? N f F
Situation 1 mg

43. 2. A car with a weight of 2500N working with a power of 130kW develops a velocity of 112 km/hour when traveling along a horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not negligible: (a) What is the car’s maximum velocity on a 50 incline hill? (b) What is the power if the car is traveling on a 100 inclined hill at 36km/h?
Situation 2 N F
Fgx f
Fgy mg