Note Chap 5 SP015 (Work, Energy and Power) Sitinorsham

Chapter 05 Physics

Exercise 5.3 :

Use gravitational acceleration, g = 9.81 m s−2

1. If it takes 4.00 J of work to stretch a spring 10.0 cm from its
initial length, determine the extra work required to stretch it an
additional 10.0 cm.

ANS. : 12.0 J

2. A book of mass 0.250 kg is placed on top of a light vertical
spring of force constant 5000 N m−1 that is compressed by 10.0
cm. If the spring is released, calculate the height of the book rise
from its initial position.

ANS. : 10.2 m

3. A 60 kg bungee jumper jumps from a bridge. She is tied to a
bungee cord that is 12 m long when unstretched and falls a total
distance of 31 m. Calculate

a. the spring constant of the bungee cord.

b. the maximum acceleration experienced by the jumper.

ANS. : 100 N m−1; 22 m s−2

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Chapter 05 Physics

4.

Figure 5.18

A 2.00 kg block is pushed against a light spring of the force

constant, k = 400 N m-1, compressing it x =0.220 m. When the

block is released, it moves along a frictionless horizontal surface

and then up a frictionless incline plane with slope  =37.0 as

shown in Figure 5.18. Calculate

a. the speed of the block as it slides along the horizontal

surface after leaves the spring.

b. the distance travelled by the block up the incline plane before

it slides back down.

ANS. : 3.11 m s−1; 0.81 m 52

Chapter 05 Physics

5. u C

A

10 m

BD

Figure 5.19

A ball of mass 0.50 kg is at point A with initial speed, u =4 m s−1
at a height of 10 m as shown in Figure 5.19 (Ignore the frictional
force). Determine

a. the total energy at point A,

b. the speed of the ball at point B where the height is 3 m,

c. the speed of the ball at point D,

d. the maximum height of point C so that the ball can pass over

it.

ANS. : 53.1 J; 12.4 m s−1; 14.6 m s−1; 10.8 m

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Chapter 05 Physics

Learning Outcome:

5.3 Power (1 hour)

At the end of this chapter, students should be able to:

 Define and use average power,

Pav = W
t

 Use instantaneous pow er,
P = F •v

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Chapter 05 Physics
5.3 Power

5.3.1 Power, P

At the end of this lesson, students should be able to:

 Define and use average power

 is a rate at which work is done.

OR a rate at which energy is transferred.

Pav = W = E
t t

 is a scalar quantity.
 S.I. unit of the power: P = W = Fs = mas
tt t
(kg)(m s−2 )(m)
unit of P =
(s)

1 hp = 746 W = kg m2 s−3 OR J s−1 OR watt (W )
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Chapter 05 Physics

At the end of this lesson, students should be able to:

 Use instantaneous power.

F

Figure 5.20 s

P = W and W = (F cos θ )s

t

P = (F cos θ )s and v = s

t t

P = Fv cosθ

OR where F : magnitude of force

 v : magnitude of velocity 
P = F •v
θ : the angle betw een F and v

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Chapter 05 Physics

Example 5.11 :

An elevator has a mass of 1.5 Mg and is carrying 15 passengers
through a height of 20 m from the ground. If the time taken to lift
the elevator to that height is 55 s. Calculate the average power
required by the motor if no energy is lost. (The average mass per
passenger is 55 kg)

Solution : h = 20 m; Δ t = 55 s

M = mass of the elevator + mass of the 15 passengers

M = 1500 + (5515) = 2325 kg

Pav = E Pav = Mgh
t t
= (2325)(9.81)(20)
55

Pav = 8294 W 57

Chapter 05 Physics

Example 5.12 :
An object of mass 2.0 kg moves at a constant speed of 5.0 m s−1
up a plane inclined at 30 to the horizontal. The constant frictional
force acting on the object is 4.0 N. Determine

a. the rate of work done against the gravitational force,

b. the rate of work done against the frictional force,

c. the power supplied to the object.

Solution : m = 2.0 kg; v = 5.0 m s −1 = constant; f = 4.0 N
N v


s

y mg sin30
x f
30 mgcos 30
30 

W = mg 58

Chapter 05 Physics

Solution : m = 2.0 kg; v = 5.0 m s −1 = constant; f = 4.0 N

a. the rate of work done against the gravitational force is given by

Wg = (mg )s cos θ and θ = 120

t t

= (mg ) s cos120 and v = s

tt

= (mg )v cos120
Wg = (2.0)(9.81)(5.0)cos120
t Wg = −49.1 W
Wg t
OR t = Fg v cos θ
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= (mg )v cos120

Wg = −49.1 W
t

Chapter 05 Physics

Solution : m = 2.0 kg; v = 5.0 m s −1 = constant; f = 4.0 N

b. The rate of work done against the frictional force is

W f = fvcos θ and θ = 180 
t

= (4.0)(5.0)cos180

W f = −20.0 W
t

c. The power supplied to the object, Psupplied
= the power lost against gravitational and frictional forces, Plost

Psupplied = Wg + W f
t t
= 49.1 + 20.0

Psupplied = 69.1 W 60

Chapter 05 Physics

Exercise 5.4 :
Use gravitational acceleration, g = 9.81 m s−2

1. A person of mass 50 kg runs 200 m up a straight road inclined
at an angle of 20 in 50 s. Neglect friction and air resistance.
Determine
a. the work done,
b. the average power of the person.

ANS. : 3.36104 J; 672 W

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Chapter 05 Physics

2.

10 1

Figure 5.21

A car of mass 1500 kg moves at a constant speed v up a road
with an inclination of 1 in 10 as shown in Figure 5.21. All
resistances against the motion of the car can be neglected. If
the engine car supplies a power of 12.5 kW, calculate the

speed v.

ANS. : 8.50 m s−1

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Chapter 05 Physics

THE END.

Next Chapter…

CHAPTER 6 :
Circular motion

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