Work, Energy, and Simple Machines Class 9 Question Answer Science Exploration Chapter 7

Students can use Exploration Class 9 Science Solutions Chapter 7 Work, Energy, and Simple Machines Question Answer NCERT Solutions as a quick reference guide.

Class 9 Science Exploration Chapter 7 Question Answer

Class 9 Science Ch 7 Work, Energy, and Simple Machines Question Answer

Work, Energy, and Simple Machines Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine (NCERT Textbook Page No. 136 – 138)

Question 1.
State whether True or False.

(i) Work is said to be done when a force is applied, even if the object does not move.
Answer:
False
Work is done only when a force is applied and the object is displaced in the direction of the force. If there is no displacement, no work is done.

(ii) Lifting a bucket vertically upward results in positive work done on the bucket.
Answer:
True
The applied force (upward) and displacement (upward) are in the same direction, so positive work is done.

(iii) The SI unit for both work and energy is joule (J).
Answer:
True
Work and energy are measured in joules (J) in SI units.

(iv) A motionless stretched rubber band has kinetic energy.
Answer:
False:
A stretched rubber band at rest has potential energy, not kinetic energy. Kinetic energy is due to motion.

(v) Energy can change from one form to another.
Answer:
True
Energy transformations are common (e.g., chemical energy in fuel → kinetic energy in a car).

Question 2.
Fill in the blanks.

(i) Work done = ____________ × ____________ (in the direction of force).
Answer:
Work done = Force × Displacement (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.
Answer:
1 joule of work is done when a force of 1 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 ____________.
Answer:
The expression for kinetic energy of a body of mass m and velocity v is ½ mv².

(iv) The potential energy of an object of mass m at a small height It from the Earth’s surface is ____________.
Answer:
The potential energy of an object of mass m at a small height h from the Earth’s surface is mgh.

(v) Power is defined as the ____________ at which work is done.
Answer:
Power is defined as the rate at which work is done.

Question 3.
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.
Answer:
(i) The force acting on the ball is zero.
False: Even at the highest point, gravity acts downward on the ball. The force is not zero.

(ii) The acceleration of the ball is zero.
False: The ball always experiences acceleration due to gravity (g = 9.8 m/s²) downward, even at the top.

(iii) Its kinetic energy is zero.
True: At the highest point, the ball’s velocity becomes zero momentarily, so its kinetic energy ½ mv² is zero.

(iv) Its potential energy is maximum.
True: At the highest point, the ball is at maximum height, so its gravitational potential energy (mgh) is maximum.

Question 4.
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 (be) a solar panel.
Answer:
Energy is never destroyed; it only transforms from one form to another depending on the situation.

(i) A truck moving uphill
Chemical energy (fuel) → Kinetic energy (motion) → Potential energy (height gained)

(ii) Unwinding of a watch spring
Elastic potential energy (spring) → Kinetic energy (movement of gears)

(iii) Photosynthesis in green leaves
Light energy (sunlight) → Chemical energy (glucose molecules)

(iv) Water flowing from a dam
Gravitational potential energy (stored water) → Kinetic energy (flowing water) → (can further convert to electrical energy in turbines)

(v) Burning of a matchstick
Chemical energy (match head) → Heat energy + Light energy

(vi) Explosion of a fire cracker
Chemical energy → Heat energy + Light energy + Sound energy + Kinetic energy of fragments

(vii) Speaking into a microphone
Sound energy (voice) → Electrical energy (microphone output)

(viii) A glowing electric bulb
Electrical energy → Light energy + Heat energy

(ix) A solar panel
Light energy (sunlight) → Electrical energy

Question 5.
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 ms^-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?
Answer:
(i) Potential energy is the energy an object has because of its position.
Formula : U = mgh
where m = mass of the object,
g = acceleration due to gravity,
h = vertical height.
When the student is lifted straight up in the elevator:
Mass m = 50 kg
Height h = 72.5 m
Gravity g = 10 m/s²
U = mgh = 50 × 10 × 72.5 = 36,250 J
So, the gain in potential energy = 36,250 joules.

(ii) When the student climbs the stairs, the vertical height reached is still the same (72.5 m).
So again:
Mass m = 5 kg,
h = 72.5 m,
g = 10 m/s²
∴ U = mgh = 50 × 10 × 72.5 = 36,250 J
The gain in potential energy is the same.

(iii) Conclusion
The path taken (straight elevator vs zig-zag staircase) does not matter.
What matters is the vertical height gained.
Potential energy depends only on m, g, h.
Hence potential energy at the top is same inspite of any path taken to reach there.

Question 6.
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.
Answer:
Crane lifting a mass

Case 1: To the 10th floor
Height = 10h (if each floor has height h)
Potential energy gained:
PE = mgh
⇒ PE₁₀ = mgh × 10 = 10 mgh

Case 2: To the 20th floor
Height = 20 h
Potential energy gained:
PE₂₀ = mg × 20h = 20 mgh

Energy comparison
∆E = PE₂₀ – PE₁₀
= 20 mgh – 10 mgh = 10 mgh
So, 10 mgh (twice of initial energy) more energy is required to lift to the 20th floor compared to the 10th floor.

Power comparison:
Power = \(\frac{\text { Work }}{\text { time }}\)
= \(\frac{\text { Energy }}{\text { time }}\)
For 10^th floor
P₁₀ = \(\frac{10 m g h}{t}\)
For 20^th floor (double time):
P₂₀ = \(\frac{20 m g h}{2 t}\)
= \(\frac{10 m g h}{t}\)
So, the power required is the same in both cases, even though more energy is needed for the 20th floor.

Question 7.
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.
Answer:
Energy required depends on:
Mass of the flag (m)
Height of the flagpole (h)
Acceleration due to gravity (g)

Formula:
Work = mgh

So, energy depends only on mass and height, not on the path or speed.
No. The work done (energy required) is the same, because the flag reaches the same height regardless of speed.
If speed is doubled, how does power change?

Power = \(\frac{\text { Work }}{\text { Time }}\)

If the flag is raised twice as fast, the time taken is halved.

P ∝ \(\frac {1}{t}\)

So, power requirement doubles, even though the total work (energy) remains the same.

Question 8.
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.
Given:
Mass of man = 60 kg
Mass of scooter = 100 kg
Mass of son = 40 kg
Velocity reached = v (same on both days)
Time taken = same on both days
Energy comes entirely from fuel.

Total mass on Day 1:
M₁ = 60 + 100= 160 kg
∵ KE = ½ mv²
then KE₁ = ½ × 160 × v² = 80v²

Total mass on Day 2:
M₂ = 60 + 40 + 100 = 200 kg
∵ KE₂ = ½ × 200 × v² = 100 v²
Since fuel energy is directly proportional
\(\frac{K E_1}{K E_2}=\frac{80 v^2}{100 v^2}\)
= \(\frac{80}{100}=\frac{4}{5}\)
Hence, the ratio of the fuel of the tank used on the two days is 4 : 5.

Question 9.
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.
Answer:
Seesaw with sliding seats
Condition for balance on a seesaw:
Torque by child = Torque by adult
Torque = Force × Distance from fulcrum.
Since Force = Weight = mg
Let child’s weight = W.
Adult’s weight = 2 W.
If child sits at distance d from fulcrum, adult must sit d at distance \(\frac { d }{ 2 }\) from the fulcrum
The adult weighs twice as much, so to balance, the adult must sit at half the distance compared to the child.

Question 10.
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 ms^-2).
Answer:
Ball thrown upward (mass = 2 kg, velocity = 20 m/s)
(i) Upward motion:
Gravity acts downward, displacement is upward → work done by gravity is negative.
Downward motion:
Gravity acts downward, displacement is downward → work done by gravity is positive.

(ii) Initial KE = ½ mv²
= ½ × 2 × (20)² = 400 J.
Without air resistance:
h = \(\frac{v^2}{(2 g)}\)
= \(\frac{(20)^2}{2 \times 10}\)
= \(\frac{400}{20}\) = 20 m.
(ball would reach 20 m).
Actual height = 19.4 m.
PE at 19.4 m = mgh
= 2 × 10 × 19.4 = 388 J.
At highest point, final KE = 0.
Total work (W_total) = change in kinetic energy (∆KE)
= ∆KE_final – ∆KE_initial
= 0 – 400 = – 400 J.
Work by gravity (W_g) = – mgh = – 388 J.
Work done by air resistance = W_total – W_g
= – 400 – (- 388) = – 12 J.

Question 11.
A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig., a variable force is applied on the block in its direction of on 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?

Answer:
(i) Here speed at 0 m.
KE = ½ mv²
⇒ 180 J = ½ × (10) × v² (∵ KE = 180 J, m = 10 kg)
⇒ 360 J= 10 v²
⇒ v = \(\sqrt{36}\) = 6 m/s

(ii) Work done by force = Area under force-displacement graph (See Fig.):
0 to 1 m (triangle, force 0 to 50 N):
Area = ½ × 1 × 50 = 25 J
1 to 3 m (rectangle, force = 50 N):
Area = 2 × 50= 100 J
3 to 4 m (triangle, force 50 to 0 N):
Area = ½ × 1 × 50 = 25 J
Total work done = 25 + 100 + 25 = 150 J
Final KE = 180 + 150 = 330 J
½ × 10 × v² = 330
⇒ v² = 66
⇒ v = \(\sqrt{66}\) ms^-1.

Negative acceleration:
The applied force is always positive (in the direction of motion) and friction is negligible, so no negative acceleration occurs in any portion.

Question 12.
The gravitational attraction on the surface of the Moon (lunar surface) is about \(\frac {1}{6}\)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?
Answer:
We know that maximum height (h) reached by an object thrown vertically with initial velocity (v) is
f = \(\frac{u^2}{2 g}\)
∴ u (initial velocity) is the same for both throws, height is inversely proportional to gravity.
⇒ h ∝ \(\frac {1}{g}\)
∴ \(\frac{h_{\text {moon }}}{h_{\text {earth }}}=\frac{\frac{1}{g_{\text {moon }}}}{\frac{1}{g_{\text {earth }}}}\)
= \(\frac{g_{\text {earth }}}{g_{\text {moon }}}\)
= \(\frac{g_{\text {earth }}}{\frac{g_{\text {earth }}}{6}}\) = 6
(∵ g_moon = \(\frac {1}{6}\) g_earth)
∴ h_moon = 6 × h_earth
= 6 × 8 = 48 m

Question 13.
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 motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig.

(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?
Answer:
(i) Clearly from the graph, the car moves at a constant speed of 35 m/s (0 acceleration) from position A to B.

(ii) Kinetic energy of car at point A = ½ mv²
= ½ × (1000) × (35)²
= 500 × 1225 = 6,12,500 J

(iii) Work done by brakes (B to C):
Final KE = 0 (car stops).
By work – energy theorem:
Work = ∆KE = 0 – 612500 = – 612500 J.
(Negative because braking force is opposite to displacement.)

(iv) Kinetic energy transfers into potential energy (and some sound) due to friction in the brakes.

Question 14.
The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. blow. At O, the velocity of the ball is 0 ms^-1 and potential energy is 30 J.
Calculate the velocity of the ball at P, Q and R.

Answer:
Point P
Potential energy is 20 J
KE = 30 J – 20 J = 10 J
Now KE = ½ mv²
10 = ½ (0.5) v²
⇒ v² = \(\frac {20}{0.5}\)
= \(\frac {200}{5}\) = 40
⇒ v² = \(\sqrt{40}\) = 2\(\sqrt{10}\) m/s

Point Q
Potential energy is 30 J
KE = 30 – 30 = 0 J
KE = ½ mv²
⇒ 0 = ½ (0.5) v²
⇒ 0 = (0.5) × v²
⇒ v = 0 m/s

Point R
Potential energy is 40 J
KE = 30 – 40 = – 10 J
– 10 = ½ (0.5) v²
⇒ \(-\frac {20}{0.5}\) = v²
⇒ v – not possible
Hence, ball cannot reach the position R. It turns back before reaching R.

Question 15.
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.
Answer:
(i) Using the conservation of energy Potential energy at top = kinetic energy at bottom
⇒ mgh = ½ mv²
⇒ v² = 2gh
⇒ v = \(\sqrt{2 gh}\)
v = \(\sqrt{2 \times 10 \times 10}\)
= \(\sqrt{200}\)
= 10\(\sqrt{2}\) m/s
Hence velocity of the coconut just before it hits the sand is 10\(\sqrt{2}\) m/s.

(ii) The work done by the resistive force of the sand equals the total energy of the coconut at the moment of impact (150 J).
Work = force × displacement
⇒ 150 = 3000 N × s
⇒ s = \(\frac {150}{3000}\)
= 0.05 m = 5 cm

Class 9 Science Chapter 2 Cell The Building Block of Life Question Answer (InText)

Think It Over (NCERT Textbook Page No. 116)

Question 1.
What will be the magnitude of velocity of the child at the bottom of the blue slide?
Answer:
When a child slides down, their potential energy at the top is converted into kinetic energy at the bottom (ignoring friction).
At the top: Potential Energy = (mgh)
At the bottom: Kinetic Energy = (½ mv²)
Equating:
mgh = ½ mv²
⇒ v² = 2 gh
⇒ v = \(\sqrt{2 gh}\)

Question 2.
Will two children of different masses reach the bottom of the same slide with the same velocity?
Answer:
Yes, they will reach the bottom with the same velocity (if friction and air resistance are negligible).
Reason: In the equation (v² = 2gh), mass does not appear.
This is a direct consequence of the principle of conservation of energy and Galileo’s observation that all bodies fall with the same acceleration due to gravity.

Question 3.
Which of the slides will result in the largest magnitude of velocity for the child at its bottom?
Answer:
The slide that will give the child the largest velocity at the bottom is the one with the greatest vertical height.
At the top of the slide, the child has gravitational potential energy = mgh.
At the bottom, this energy is converted into
kinetic energy = ½ mv²
Equating: ½ mv² = mgh
We know that the velocity depends only on height (h) and gravity (g), not on the mass of the child or the shape of the slide.

Conclusion:
The slide with the maximum vertical height will result in the largest magnitude of velocity for the child at its bottom.

Pause and Ponder (NCERT Textbook Page No. 119)

Question 1.
In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig.). Is she doing any work on the barbell while holding it steady?

Answer:
No, the weightlifter is doing zero work on the barbell while holding it steady, as work is defined as force multiplied by displacement (W = fs). Since the barbell is not moving, the displacement (s) is zero, regardless of the force applied.

Question 2.
Is the work done by friction on the stack of coins that travels on a rough surface (Fig.) positive, negative or zero?

Answer:
The work done by friction on the sliding stack of coins is negative. Friction opposes the motion, acting in the opposite direction of the displacement, which removes kinetic energy from the system.

Pause and Ponder (NCERT Textbook Page No. 121)

Question 3.
When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride? (Page 121)
Answer:
When we pedal a bicycle on a flat road, the chemical energy from our muscles is transformed into several different forms:

Kinetic Energy:
This is the primary form. Our muscles move the pedals, which turns the wheels and puts the bicycle and our body in motion.

Heat (Thermal Energy):
This happens in two ways. First, our body generates heat as a byproduct of muscle contraction. Second, friction in the bike’s moving parts (like the chain and bearings) and the tires rubbing against the road creates heat.

Sound Energy:
You might hear the whirring of the chain or the tires on the pavement, which is a small portion of energy being converted into sound waves.

Pause and Ponder (NCERT Textbook Page No. 123)

Question 4.
Two objects A and B of mass m and 4 m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B? (Page 123)
Answer:
We know kinetic energy (KE) = ½ mv²
where m is mass and v is velocity.
Now, m_A = m
and m_B = 4 m
and KE_A = KE_B
⇒ ½ m_Av_A² = ½ m_Bv_B²
⇒ mv_A² = mv_B²
⇒ v_A² = v_B²
⇒ v_A = 2v_B
⇒ \(\frac{v_A}{v_B}=\frac{2}{1}\) = 2 : 1

Question 5.
Does the kinetic energy of an object which moves with constant velocity change with its position?
Answer:
No, the kinetic energy of an object moving with constant velocity does not change with its position.

Velocity Dependency:
Kinetic energy depends only on the mass and speed (magnitude of velocity) of an object.

Constant Factors:
If the mass is constant and the velocity is constant, the kinetic energy (1/2 mv2) remains unchanged regardless of the object’s location on a flat path.

Pause and Ponder (NCERT Textbook Page No. 126)

Question 6.
Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?
Answer:
No, the potential energy does not change when an object moves horizontally at a constant velocity. This is because its height relative to the ground remains the same, and potential energy (p = mgh) only depends on vertical position.

However, if the object is gradually raised in the vertical direction, its potential energy increases. As height increases, the work done against gravity is stored as additional gravitational potential energy.

Pause and Ponder (NCERT Textbook Page No. 129)

Question 7.
For the situation depicted in Fig., calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh. (Page 129)

Answer:
A ball of mass m is dropped from a height h.
Now mechanical energy = kinetic energy + potential energy
M = KE + PE
At the ground level, potential energy is zero.
PE = 0
The ball has gained speed due to gravity. Its velocity just before impact can be found using energy conservation law.
∴ v² = u² + 2gh, Here u = 0
v² = 2gh
1 , 1
∴ K = ½ mv²
= ½ m (2gh) = mgh

Question 8.
You may have seen an exhibit like that in Fig. in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?

Answer:
Point A (highest point):
The ball has maximum potential energy = mgh.
The kinetic energy is zero because the ball is momentarily at rest before release.

Point B (midway down):
The ball has fallen partway, so its potential energy decreases.
Its kinetic energy increases because it is moving faster.
The sum of KE + PE remains constant (equal to the initial mgh).
Point B (lowest point near the ground).
The potential energy is nearly zero. (since height ≈ 0).
The kinetic energy is maximum.
Total mechanical energy is still mgh.
Subsequent points (C, D, E…) have lower heights because in a real-world exhibit, the ball doesn’t rise back to the same height after each swing. Instead, each successive peak is lower.

This happens because:
Friction at the pivot (where the ball is attached) converts some mechanical energy into heat.
Air resistance dissipates energy as the ball moves through the air.
Sound energy may also be produced when the ball moves or collides.
So yes the lower heights are directly due to energy lost to non-conservative forces like friction and air resistance. Mechanical energy is not perfectly conserved in practice, which is why the ball eventually comes to rest.

Pause and Ponder (NCERT Textbook Page No. 132)

Question 9.
Explain, why roads on hills are built to wind around in gentle slopes rather than going straight up:

Answer:
Reduced Effort (Lower Gradient):
Roads on hills wind around (zig-zag) to increase the distance over which elevation is gained, creating a gentle slope (smaller angle of inclination). A straight road would be very steep, requiring a much higher engine force to move a vehicle up, which is harder for vehicles.

Safety (Increased Friction):
A gentle slope increases the surface area and angle of traction, reducing the risk of vehicles skidding or slipping backward. Steep, straight roads have less frictional force acting on the tires, making them prone to accidents.

Preventing Engine Strain:
A gentle, winding path helps in keeping the engine load lower, preventing overheating and reducing brake failure during descent.

Question 10.
To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig.). Explain why

Answer:
Reduced Effort (Mechanical Advantage):
An inclined ladder acts as an inclined plane (a simple machine). While the work to reach a specific height is the same, the effort required is spread over a longer distance, meaning less force is needed at any given moment.

Reduced Effect of Gravity:
When climbing vertically, we are directly lifting our full body weight against gravity. On an incline, the slope supports part of our weight, reducing the direct vertical force of gravity we have to fight.

Better Stability/Traction:
It is easier to maintain balance and get a secure foothold/handhold on a slanting surface compared to a vertical one, which requires more strength to keep from pulling away from the wall.

Pause and Ponder (NCERT Textbook Page No. 135)

Question 11.
Why is it easier to open the lid of a can by using a spoon as shown in Fig. ?

Answer:
Using a spoon works as a first-class lever. The rim of the can acts as a fulcrum (pivot point). Because the handle of the spoon is much longer than the tip under the lid, it increases the distance from the pivot point (effort arm). According to the principle of leverage, a greater distance allows a smaller, easier force (input effort) to produce a much higher lifting force (output force) on the lid.

Question 12.
Why do you push an object closer to scissors (fulcrum) when you want to cut an object which is hard?
Answer:
Scissors are double-levers, with the screw in the middle acting as the fulcrum. The force you apply with your hand is amplified when the item being cut is closer to this fulcrum. Placing the object closer to the pivot point reduces the load arm (distance from the object to the pivot), maximizing the cutting force because less effort is lost to leverage.

Question 13.
Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.
Answer:
All real machines involve friction between moving parts. Friction converts mechanical energy into thermal energy (heat). According to the work- energy theorem and conservation of energy, the total energy remains constant – but friction continuously converts useful mechanical energy into thermal energy, which cannot be automatically recovered as mechanical energy. As a result, the mechanical energy of the machine decreases over time and it eventually slows down and stops. A perpetual motion machine would require zero energy loss, which is impossible in reality. Machines only transform energy; they cannot create it.