Class 9 · ScienceNCERT 2026 Revised Syllabus
Chapter 7: Work, Energy, and Simple Machines (0906CH07)
Exploration · Work, kinetic and potential energy, power, simple machines.
Chapter code: 0906ch07 · Book: Exploration · Class 9 Science
Find complete NCERT solutions for Class 9 Science, Exploration Chapter 7. Every question has a clear, step-by-step answer for chapter code 0906CH07.
Pause and Ponder(13 questions)
1
In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?
School Level Solution
Ans.
Mechanical work is W = F × displacement in the direction of force. While the barbell is held steady its displacement is zero, so the work done on the barbell is 0.
2
Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c) — positive, negative or zero?
School Level Solution
Ans.
Friction acts opposite to the displacement of the moving stack, so the work done by friction is negative.
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?
School Level Solution
Ans.
The chemical energy supplied by the muscles is converted mainly into kinetic energy of the bicycle and rider. Part is dissipated as thermal energy because of friction and air resistance, and a small part appears as sound and other losses.
4
Two objects A and B of mass m and 4m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B?
School Level Solution
Ans.
Since K = ½mv² and the kinetic energies are equal, m vA2 = 4m vB2. Hence vA : vB = 2 : 1.
5
Does the kinetic energy of an object which moves with constant velocity change with its position?
School Level Solution
Ans.
Kinetic energy is K = ½mv². If mass and velocity are constant, kinetic energy does not change with position.
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?
School Level Solution
Ans.
Near Earth’s surface PE = mgh. Horizontal motion at constant height leaves h unchanged, so PE does not change. Raising the object increases h, so its potential energy increases.
7
For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.
School Level Solution
Ans.
Mechanical energy = kinetic + potential. Just before impact, the ball has converted the initial potential energy into kinetic energy (with zero height reference at the ground), so E = K + U = mgh.
8
You may have seen an exhibit like that in Fig. 7.22 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?
School Level Solution
Ans.
At a high point A the ball has more potential energy and less kinetic energy. Moving downward toward B, PE decreases and KE increases. Climbing toward C, KE decreases and PE increases. In a real system, friction and air resistance convert part of the mechanical energy into thermal energy, so successive turning points are lower.
9
Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?
School Level Solution
Ans.
For a given height, a gentler slope means a longer path. The component of the vehicle’s weight along the slope is smaller, so the required driving force is reduced. The total gain in gravitational potential energy is still mgh.
10
To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.
School Level Solution
Ans.
An inclined ladder increases the distance over which the same vertical height is gained, reducing the component of gravitational force that must be overcome at any instant. The work against gravity remains approximately mgh, but the required force is smaller.
11
Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?
School Level Solution
Ans.
The spoon is a lever. Its long effort arm produces a larger turning effect (torque) for the same hand force, making it easier to lift the lid.
12
Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?
School Level Solution
Ans.
For a lever, torque balance gives effort × effort arm = load × load arm. Bringing the hard object closer to the scissors’ fulcrum decreases the load arm, so the same effort produces a larger force at the cutting point.
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.
School Level Solution
Ans.
Real machines cannot avoid dissipative forces such as friction and air resistance. These convert useful mechanical energy into thermal energy (and sound, deformation, etc.). Without continuous external energy input, the machine’s useful mechanical energy decreases and it eventually slows and stops.
Chapter Exercises(6 questions)
1
A weightlifter holds a barbell steady in her hands. Is she doing any work on the barbell while holding it steady?
School Level Solution
Ans.
Work = force × displacement in the force’s direction. The barbell does not move, so displacement is zero → work done is zero, however tired her muscles feel (that energy goes into body heat, not the barbell).
2
Is the work done by friction on the sliding coin-stack — positive, negative or zero?
School Level Solution
Ans.
Friction acts opposite to the stack’s motion, so the angle between force and displacement is 180° → work = F·s·cos180° = negative. That is why friction removes kinetic energy and slows the stack.
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?
School Level Solution
Ans.
Muscular (chemical) energy becomes: kinetic energy of rider + cycle; heat in muscles and from tyre/chain friction; energy lost to air resistance and a little sound. At steady speed the input exactly balances these losses.
4
Two objects A (mass m) and B (mass 4m) have the same kinetic energy. What is the ratio of the magnitudes of their velocities?
School Level Solution
Ans.
½mv_A² = ½(4m)v_B² → v_A² = 4v_B² → v_A = 2v_B. So the lighter object must move twice as fast; ratio v_A : v_B = 2 : 1.
5
Does the kinetic energy of an object moving with constant velocity change with its position?
School Level Solution
Ans.
KE = ½mv²: with constant velocity, v never changes, so KE stays identical at every position. (Potential energy may still vary with height — that is a separate store.)
6
Does the potential energy of an object near Earth’s surface change moving horizontally at constant velocity? What if it is gradually raised vertically?
School Level Solution
Ans.
Gravitational PE = mgh depends only on height. Moving horizontally, h is fixed → PE unchanged. Raised vertically, h grows → PE increases, the extra energy coming from the work done in lifting.
Activities — How to Do(3 questions)
1
Activity 7.2 (pendulum energy): release a pendulum bob from the drawn horizontal level and watch the first swings. Does it nearly regain that level? How do kinetic and potential energy change at points P, Q and R?
School Level Solution
Ans.
At release point P: max PE, zero KE. At bottom Q: PE → max KE (fastest). At far end R: KE → PE, nearly regaining the line. The bob almost reaches it — air friction steals a little energy each swing.
2
Activity 7.3 (cart up a ramp): lift a cart vertically with a spring balance, then pull it up planks of decreasing steepness to the same height. How does the required force change?
School Level Solution
Ans.
Vertical lift needs the full weight. Up a plank, the balance reads less — and still less as the plank flattens. Work stays mgh in all cases (ignoring friction): a gentler slope trades longer distance for smaller force — the inclined plane’s advantage.
3
Activity 7.5 (beam balance): hang a scale-beam with coin pans; balance 1 coin against 2, then 4, then 8 coins by shifting the heavier pan inward. What relation do the readings verify?
School Level Solution
Ans.
With 1 coin left and 2 right, balance returns when the right pan sits at half the left distance. Repeating with 4 and 8 coins confirms effort × effort-arm = load × load-arm (n₁L₁ = n₂L₂) — moments balance, the lever rule behind beam scales.