1. WORK
- Is the amount of force applied to
an object over a distance

2. 3 CONDITIONS FOR WORK TO BE DONE
- There must be a force acting on the
object.
- The object has to move a certain
distance called the displacement.
- There must be a component of the
force in the direction of the motion.

3. WORK IS NOT DONE WHEN :
- The object is stationary
- No force applied on the object in the
direction of the motion
- The direction of the motion of the object
is perpendicular to that of the applied
force
- When work is done and energy is
transferred to the object

4. A B

5. A B

6. A boy applies a force to a wall and becomes
exhausted.
Fd

7. A waiter carries a tray full of meals by his arm
across the room
F
d

8. CALCULATING THE AMOUNT OF WORK
( F cos θ ) dW =
d – displacement | meters (m)
F – force parallel to the displacement | newton (N)
Θ – the angel between the force and the
displacement
W – work | newton-meters or joules ( J )

9. A book weighing 1.0 N is lifted 2 m. How much
work was done?
It took 50 J to push a chair 5 m across the
floor. With what force was the chair pushed?
A force of 100 N was necessary to lift a rock.
A total of 150 J of work was done. How far was
the rock lifted?
- 2 J
- 10 N
- 0.666 m

10. DID YOU KNOW ?
If you have to lift a new sofa to a second floor
apartment, the work done against the gravity is
the same whether you haul it straight up the
side of the building with ropes or take a longer
path up the stairs.
Only the vertical distance matters because the
force of gravity is vertical

11. A box is dragged across a
floor by a 100N force directed
60o above the horizontal. How
much work does the force do in
pulling the object 8m?
Try to solve:

12. A porter pulls a 10kg luggage along a
level road for 5 m by exerting a force of 20N
at an angle of 30o with the horizontal
shoulder through a vertical distance of 1.5
meters and carries it for another 5 meters.
How much work does he do in
a.) pulling
b.) lifting
c.) carrying the luggage on his shoulder
Solve:

13. a.) pulling the luggage
Given: F – 20 N Find: W
Θ – 30o
d – 5 m
Sol’n : W = ( F cos θ ) d
= (20 N) (cos 30o) (5m)
W = 87 J

14. b.) lifting the luggage
Given : m – 10 kg Find: W
d – 1.5 m
Sol’n : W = F x d
F = ?
= (98 N) (1.5m)
W = 147 J

15. The force on the luggage is perpendicular to the
direction of motion. The distance moved in the
direction of force is zero. Therefore, work is
zero.
Hence, the porter does no work in carrying the
luggage.
c.) carrying the luggage
W = O

16. ENERGY
- The capacity to do work

17. - the energy possessed by bodies in
motion
Kinetic Energy
Potential Energy
-stored energy
- Associated with forces that depend on
the position or configuration of a body
and its surroundings.

22. POTENTIAL ENERGY: THE STORED ENERGY
|Gravitational Potential Energy|
- the energy of an object due to its
higher position in the gravitational
field
PEg = mgh

23. Solve :
A 800g ball is pulled up a slope
as shown in the diagram. Calculate
the potential energy it gains.
20 cm
50 cm

24. A box has a mass of 5.8kg.
The box is lifted from the garage
floor and placed on a shelf. If the
box gains 145J, how high is the
shelf?
Solve :

25. |Elastic Potential Energy|
- stored energy in a spring by
stretching or compressing it
Pes =
1
2
𝑘𝑥2

26. k – spring constant
x – stretched distance of the elastic object
k x
F

27. When a 4 kg mass is hung vertically on a
certain light spring that obeys Hooke's
law, the spring stretches 2.5 cm. If the 4
kg mass is removed,
(a)how far will the spring stretch if a 1.5
kg mass is hung on it, and
(b)(b) how much work must an external
agent do to stretch the same spring 4
cm from its unstretched position?

28. (a)We find the spring constant of
the spring from the given data.
F = kx.
F = mg = -(4 kg)(9.8 m/s2) = -39.2N.
k = F/x = (39.2 N)/(0.025 m) = 1568
N/m.

29. Now we use x = F/k to find the
displacement of a 1.5 kg mass.
F = (1.5kg)(9.8m.s2)
= 14.7 N.
x = (14.7 N)/(1568 N/m)
= 0.009375 m
= 0.975 cm.

30. (b) W = (1/2)kx2
= (1/2)(1568 N/m)(0.04m)2
= 1.2544 Nm
= 1.2544 J

31. A 5 cm stretched spring with a 15 g
bag of salt hanging on it is lifted up
by 7 meters.
Calculate its elastic potential energy .

32. KINETIC ENERGY: ENERGY IN MOTION
KE =
1
2
𝑚𝑣2
m - mass of the object
v– velocity

33. Solve :
calculate the kinetic energy of
a 1000 kg car travelling at 60 km/h.
KE =
1
2
𝑚𝑣2
=
1
2
(1000kg)(60𝑘𝑚/ℎ)2
= 138 944.45 J

34. A van has a mass of 3000 kg and
a car has a mass of 1500 kg.
if both are travelling at the same
speed, how would you compare
their kinetic energy?
Solve :

35. Van : KE = ½ (3000 kg) 𝑣2
KE = 1500 𝑣2
Car : KE = ½ (1500 kg) 𝑣2
KE = 750 𝑣2
The van’s KE is twice the KE of the car.

36. POWER
- the rate of doing work

37. CALCULATING POWER
P =
𝑊
𝑡
W = work done | newton meters (J)
t = time | seconds (s)
P = power | joules per second (J/s)
or watts (W)

38. Solve :
Dan climbs a flight of stairs in
1.5 minutes. If he weighs 450 N and
the stairs is 10 meters from the
ground, how much power will he
develop?
Given : t = 1.5 min / 90 s
F = 450 N
d = 10 m

39. P =
𝑊
𝑡
=
(450 𝑁)(10 𝑚)
90 𝑠
= 50 N ● m/s
or 50 W

40. DID YOU KNOW ?
The unit of power, watt (W),
was named after the Scottish
mathematician and engineer –
JAMES WATT

41. Solve :
Andy and Bryan each lift a 150 kg barbell at
a height of 1.5 m off the ground. Andy lifts
his barbell in 1 second and Bryan lifts his
barbell in 2 seconds.
a.) who does more work ? Explain.
b.) who exerts more power ?

42. a.) both do the same amount of work.
b.) Andy : P = W/t
P = (150 kg) (9.8 m/s2) (1.5 m)
1 s
P ≈ 2200 watt
Bryan : P = W/t
P = (150 kg) (9.8 m/s2) (1.5 m)
2 s
P ≈ 1100 watt

43. SIMPLE MACHINES