Work, Energy, and Simple Machines - Chapter 7 Class 9 Exploration

State whether True or False.
(i) Work is said to be done when a force is applied, even if the object does not move.
(ii) Lifting a bucket vertically upward results in positive work done on the bucket.
(iii) The SI unit for both work and energy is joule (J).
(iv) A motionless stretched rubber band has kinetic energy.
(v) Energy can change from one form to another.
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(i)
False
— work needs a displacement, not just a force.
(ii)
True
— force (up) and displacement (up) are in the same direction.
(iii)
True
— both are measured in joules.
(iv)
False
— a motionless stretched band has
potential
energy, not kinetic.
(v)
True
— energy can change from one form to another.
Each statement checked
Statement
True / False
Why
(i) work is done even if the object does not move
False
W = F × s. If s = 0, then W = 0 — a force alone is not work.
(ii) lifting a bucket up → positive work
True
Force (up) and displacement (up) are in the same direction.
(iii) SI unit of work and energy is the joule
True
Both are measured in joule (J); 1 J = 1 N × 1 m.
(iv) a motionless stretched rubber band has kinetic energy
False
It is not moving — it stores
potential
energy, not kinetic.
(v) energy can change from one form to another
True
Electrical → light, chemical → mechanical, and so on.
← Back to: 7.1 Work Done by a Constant Force

Fill in the blanks.
(i) Work done = ______ × ______ (in the direction of force).
(ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force.
(iii) The expression for kinetic energy of a body of mass m and velocity v is ______.
(iv) The potential energy of an object of mass m at a small height h from the Earth's surface is ______.
(v) Power is defined as the ______ at which work is done.
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(i)
force
×
displacement
.
(ii)
1
newton.
(iii)
K =
1
2
mv
2
.
(iv)
U = mgh
.
(v)
rate
at which work is done.
The five blanks filled
Blank

When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?
(i) The force acting on the ball is zero.
(ii) The acceleration of the ball is zero.
(iii) Its kinetic energy is zero.
(iv) Its potential energy is maximum.
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Correct:
(iii) and (iv)
.
At the highest point the ball is momentarily at rest, so its kinetic energy is zero (iii) and its potential energy is maximum (iv). The force of gravity still acts, so the force is not zero and the acceleration is g (not zero) — (i) and (ii) are wrong.
At the highest point — what is TRUE
What is FALSE
(iii) Kinetic energy is zero
— the ball has stopped rising, v = 0
(i) The force is zero —
no
, gravity still acts (mg)
(iv) Potential energy is maximum
— it is at its greatest height
(ii) The acceleration is zero —
no
, a = g, still 10 m s⁻² downwards
All the KE has turned into PE
Zero velocity does
not
mean zero force or zero acceleration
⇒ correct options are
(iii)
and
(iv)
—
← Back to: 7.4.3 Conservation of mechanical energy

For each of the following situations, identify the energy transformation that takes place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, and (ix) a solar panel.
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(i) Chemical (fuel) → kinetic + gravitational potential energy.
(ii) Elastic potential → kinetic (mechanical) energy.
(iii) Light (solar) → chemical energy.
(iv) Gravitational potential → kinetic energy.
(v) Chemical → heat + light energy.
(vi) Chemical → heat + light + sound + kinetic energy.
(vii) Sound → electrical energy.
(viii) Electrical → light + heat energy.
(ix) Light (solar) → electrical energy.
Energy transformation in each situation
Situation
Energy transformation
(i) a truck moving uphill
chemical (fuel) → kinetic + potential (+ heat)
(ii) unwinding of a watch spring
elastic potential → kinetic
(iii) photosynthesis in green leaves
light (solar) → chemical
(iv) water flowing from a dam
potential → kinetic
(v) burning of a matchstick
chemical → heat + light
(vi) explosion of a fire cracker
chemical → heat + light + sound + kinetic
(vii) speaking into a microphone
sound → electrical
(viii) a glowing electric bulb
electrical → light + heat
(ix) a solar panel
light (solar) → electrical
← Back to: 7.3 Forms of Energy

A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s⁻², and student's mass is m = 50 kg. (i) Find the gain in the potential energy if the student is lifted straight up to the top. (ii) Find the gain in the potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about the dependence of the potential energy on the path taken?
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(i)
U
= mgh
= 50 × 10 × 72.5
= 36250
J
.
(ii) Climbing the stairs reaches the same height, so the gain is the
same
:
36250
J
.
(iii) The gain in potential energy depends only on the
vertical height
, not on the path taken.
Gain in potential energy — lift vs stairs
(i) Lifted straight up in the elevator
U
= mgh
= 50 × 10 × 72.5
=
36250 J
↓
(ii) Climbing the staircase to the same top
the height gained is the
same
72.5 m →
U =
36250 J
↓
Compare the two
the path was long and winding, but
h
did not change
↓
(iii) The gain in potential energy depends only on the
height
,
not on the path
taken
← Back to: 7.4.2 Potential energy

A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
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Let the height to the 10th floor be
h
(so the 20th floor is
2h
) and the first time be
t
.
Energy:
E
1
= mgh
;
E
2
= mg(2h) = 2mgh
— so
twice
the energy (an extra
mgh
).
Power:
P
1
=
mgh
t
;
P
2
=
2mgh
2t
=
mgh
t
— the power is the
same
.
Energy and power for the crane
First lift — to the 10th floor
height
h
, time
t
→ energy
E = mgh
, power
P = mgh/t
↓
Second lift — to the 20th floor
height is
2h
, so energy
= mg(2h) =
2E
—
twice
the energy
↓
But it takes double the time
P'
=
2mgh
2t
=
mgh
t
↓
Twice
the energy, but the
same
power
← Back to: 7.5 Power

Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
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The energy (work) needed depends on the
weight of the flag
and the
height of the pole
:
W = mgh
.
Raising it slowly or quickly does
not
change the work done (work depends on force and distance, not time).
If the speed is doubled, the time halves, so power
P =
W
t
doubles
.
What does NOT change
What DOES change
Energy to raise the flag =
mgh
— it depends only on the mass of the flag and the height of the pole
Power
= work ÷ time
Raising it slowly or quickly does
not
change the work done
Raise it in half the time and you need
twice
the power
Work is the same either way
So if the speed is
doubled
, the power requirement is
doubled
← Back to: 7.5 Power

A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
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Day 1 mass
= 60 + 100 = 160
kg
; Day 2 mass
= 60 + 100 + 40 = 200
kg
.
Fuel
∝
kinetic energy
=
1
2
mv
2
, and
v
is the same, so fuel
∝
mass.
Ratio (Day 1 : Day 2)
= 160 : 200 =
4 : 5
.
Fuel used on the two days
Day 1 — man + scooter
m = 60 + 100 =
160 kg
→
E
1
=
1
2
(160)v
2
↓
Day 2 — man + son + scooter
m = 60 + 40 + 100 =
200 kg
→
E
2
=
1
2
(200)v
2
↓
Same speed
v
, same time → the fuel is proportional to the energy
E
1
E
2
=
160
200
↓
Fuel ratio =
4 : 5
← Back to: 7.4.1 Kinetic energy

On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
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For balance: weight × distance must be equal on both sides. The adult's weight is twice the child's, so the adult must sit at
half the distance
from the fulcrum that the child sits.
For example, if the child sits 2 m from the fulcrum, the adult sits 1 m from the fulcrum (since
W × 2 = 2W × 1
). The child sits farther; the heavier adult sits closer.
Balancing the seesaw
The lever rule
n
1
L
1
= n
2
L
2
— effort × effort arm = load × load arm
↓
The adult weighs
twice
the child
so the adult's weight is 2W and the child's is W
↓
Put it in the rule
W × L
child
= 2W × L
adult
↓
The adult must sit at
half
the child's distance from the fulcrum
← Back to: 7.6.3 Lever

A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻¹. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s⁻²).
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(i) Upward motion: gravity (down) opposes displacement (up) →
negative
work. Downward motion: gravity (down) is along displacement (down) →
positive
work.
(ii) Initial KE
=
1
2
× 2 × 20
2
= 400
J
. At the top, PE
= mgh
= 2 × 10 × 19.4
= 388
J
, KE = 0.
Energy lost to air resistance
= 400 - 388 = 12
J
, so the work done by air resistance
=
-12 J
.
Ball thrown up at 20 m s⁻¹
(i) Going
up
gravity acts down, displacement is up → work by gravity is
negative
↓
(i) Coming
down
gravity acts down, displacement is down → work by gravity is
positive
↓
(ii) Without air, the ball would rise to
h
=
v
2
2g
=
400
20
= 20
m
↓
It only reaches 19.4 m
the missing energy went into work against the air
↓
W
air
= mg(19.4 - 20)
= 2 × 10 × (-0.6)
=
-12 J
← Back to: 7.1.2 Positive and negative work done

A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block's speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?
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(i) At 0 m:
1
2
mv
2
= 180
⇒ v
2
=
360
10
= 36
⇒ v
= 6
m s
-1
.
Work by the force = area under Fig. 7.37
=
1
2
(1)(50) + (2)(50) +
1
2
(1)(50)
= 25 + 100 + 25
= 150
J
.
(ii) KE at 4 m
= 180 + 150
= 330
J
⇒ v
2
=
660
10
= 66
⇒ v
=
66
≈ 8.1
m s
-1
.
The applied force stays in the direction of motion (always positive) throughout, so the block keeps speeding up — it has
no negative acceleration
in any portion.
Reading work off the force-displacement graph
Work done =
area
under the F–s graph
from 0 m to 4 m: a trapezium of height 50 N
↓
Area
1
2
(4 + 2) × 50
= 150
J
↓
(i) Speed at 0 m
180
=
1
2
(10)v
2
⇒ v
=
6 m s
-1
↓
(ii) Speed at 4 m
KE = 180 + 150 = 330 J
⇒ v
=
66
≈
8.1 m s
-1
↓
The force is always
along
the motion, so the work is positive throughout —
no negative acceleration anywhere
← Back to: 7.4.1 Kinetic energy

The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6 th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
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For the same throwing speed,
1
2
mv
2
= mgh
⇒ h
=
v
2
2g
, so
h ∝
1
g
.
On the Moon
g
is
1
6
of Earth's, so the height is
6 times
larger:
h
moon
= 6 × 8
= 48
m
.
The same throw on the Moon
The astronaut throws with the same velocity
so the kinetic energy
1
2
mv
2
is the
same
↓
All of it turns into potential energy
1
2
mv
2
= mgh
⇒ h
=
v
2
2g
↓
On the Moon g is
one-sixth
so
h
becomes
six times
larger
↓
h
Moon
= 6 × 8
=
48 m
← Back to: 7.4.2 Potential energy

A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car (Fig. 7.38) starts from the instant the driver spots the traffic. (i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?
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(i) Between A and B the car moves at
constant speed
(35 m s⁻¹) — this is the driver's reaction time before braking.
(ii) KE at A
=
1
2
× 1000 × 35
2
= 612500
J
.
(iii) The brakes bring the car from 35 m s⁻¹ to rest, so work by the brakes
= 0 - 612500 =
-612500 J
.
(iv) The kinetic energy transforms mainly into
heat
(in the brakes, tyres and road) and some
sound
.
The braking car
(i) Between A and B
the speed-time graph is
flat
at 35 m s⁻¹ — the car moves at a
constant speed
(the driver's reaction time)
↓
(ii) Kinetic energy at A
K
=
1
2
(1000)(35)
2
=
612500 J
↓
(iii) Work done by the brakes, B → C
the car stops, so
Δ K = 0 - 612500
→
W =
-612500 J
↓
(iv) The kinetic energy is transformed into
heat
(and some sound) in the brakes, tyres and road
← Back to: 7.4.1 Kinetic energy

The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s⁻¹ and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
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Frictionless, so total mechanical energy = value at O =
30 J
(all potential there). At any point,
KE = 30 - PE
and
v
=
2 KE
m
=
4 KE
(since m = 0.5 kg).
At P (PE = 20 J):
KE
= 10
J
⇒ v
=
40
≈ 6.3
m s
-1
.
At Q (PE = 30 J):
KE
= 0
⇒ v
= 0
m s
-1
.
At R (PE = 40 J): this needs 40 J but only 30 J is available, so the ball
cannot reach R
— it turns back before R. (Read the exact PE values from your copy of Fig. 7.39.)
Velocity from the potential energy-displacement graph (total energy = 30 J)
Point
Potential energy
Kinetic energy = 30 − U
v =
2K/m
, m = 0.5 kg
O
30 J
0 J
0 m s⁻¹
(given)
P
20 J
10 J
2 × 10 / 0.5
=
6.3 m s
-1
Q
30 J
0 J
0 m s⁻¹
R
40 J
—
cannot be reached — U > total energy, so the ball never gets to R
← Back to: 7.4.3 Conservation of mechanical energy

A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut's energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s⁻².
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(i)
v
=
2gh
=
2 × 10 × 10
=
200
≈ 14.1
m s
-1
.
(ii) Energy at impact
= mgh
= 1.5 × 10 × 10
= 150
J
. This equals work done against the sand
= F × d
:
3000 × d
= 150
⇒ d
= 0.05
m
= 5
cm
.
The falling coconut
(i) Falling 10 m from the tree
all the PE becomes KE:
1
2
mv
2
= mgh
↓
Solve for v
v
=
2gh
=
2 × 10 × 10
=
14.1 m s
-1
↓
(ii) All that energy makes the depression
energy
= mgh
= 1.5 × 10 × 10
= 150
J
↓
The sand resists with 3000 N over a depth d
3000 × d = 150
↓
Depth of the depression
d = 0.05 m = 5 cm
← Back to: 7.4.3 Conservation of mechanical energy

Project 1
Remove both the ends from a pen so that the refill can slide freely through the barrel (Fig. 7.40). Fix the pen cap to the side of the barrel and attach a rubber band to the clip of the cap. Connect the free end of the rubber band to the refill using a safety pin. Stretch and release the rubber band. The refill shoots out, showing the conversion of elastic potential energy into kinetic energy. Repeat with different amounts of stretch, and observe how the distance travelled changes. Is there a relationship between the stretch and the distance travelled?
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The more you stretch the rubber band, the more elastic potential energy it stores, so the refill leaves faster and travels farther. You will find that a greater stretch gives a greater launch distance — the stored potential energy converts into more kinetic energy.

Project 2
Construct one or more simple machines, or a combination of them (lever, pulley, and inclined plane) using easily available materials, such as cardboard, wooden strips or rulers, pencils or bolts (to act as a fulcrum), thread or rope, small pulleys (or two bottle caps stuck together), and paper cups to hold small weights. Be imaginative in your design. Use your model to lift or move a small load, measure the effort and the load, and calculate the mechanical advantage.
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Build, say, a lever (ruler on a pencil) or a pulley (bottle caps) and lift a cup of coins. Measure the load (weight lifted) and the effort (force you apply), then compute mechanical advantage = load ÷ effort. A model with a longer effort arm or a longer ramp should give a mechanical advantage greater than 1.

Project 3
Computer simulations can help in visualising physical quantities that are difficult to observe directly. The PhET simulations (https://phet.colorado.edu) provide interactive models, such as Energy Skate Park, Energy Forms and Changes, Pendulum Lab, and Masses and Springs. Use these to explore how different forms of energy change as parameters, such as mass, height, and friction are varied.
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In Energy Skate Park, watch the KE and PE bars swap as the skater rises and falls while the total stays constant — until you add friction, when the total mechanical energy slowly drops as thermal energy rises. Changing mass and height scales the energies, showing K = ½mv² and U = mgh in action.
🌌 The Quest Continues
Scientists have found that the Universe is not only expanding but expanding
faster and faster
. To explain this, they propose a mysterious form of energy called
dark energy
, which cannot be exchanged with other forms of energy.
Studying its effects matters because dark energy may decide the fate of the Universe billions of years into the future.