RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Forces that change direction and magnitude are awkward to handle with F = ma alone, and that is where energy methods pay off. You will define work as a scalar product, connect it to kinetic energy through the work-energy theorem, and use conservation of mechanical energy to solve problems in one line that would otherwise take a page. Collisions then show what happens when energy is not conserved but momentum still is.
A satellite moves in a circular orbit around the Earth at constant speed. The work done by the gravitational force on it in one complete revolution is:
Answer
Zero — the gravitational force is directed towards the centre while the velocity is tangential, so the angle between force and displacement is 90° at every instant and W = Fs cos 90° = 0. Independently, gravity is conservative, so the work done over any closed path is zero.
A body of mass 2 kg has a momentum of 10 kg m/s. What is its kinetic energy?
Answer
25 J — using K = p²/2m = (10)²/(2 × 2) = 100/4 = 25 J. Alternatively v = p/m = 10/2 = 5 m/s, so K = ½(2)(25) = 25 J.
A spring of force constant 200 N/m is stretched from its natural length by 10 cm. The work done in stretching it is:
Answer
1 J — the work stored as elastic potential energy is W = ½kx² = ½ × 200 × (0.10)² = 100 × 0.01 = 1 J. The common error is to use W = kx² or to leave x in centimetres.
Two bodies of equal mass undergo a perfectly elastic head-on collision, one moving and the other at rest. After the collision:
Answer
They exchange velocities — for a one-dimensional elastic collision with m₁ = m₂, the standard results v₁ = [(m₁ − m₂)u₁ + 2m₂u₂]/(m₁ + m₂) and v₂ = [(m₂ − m₁)u₂ + 2m₁u₁]/(m₁ + m₂) reduce to v₁ = u₂ and v₂ = u₁. So the incoming body stops dead and the struck body moves off with the incoming speed, as seen with equal-mass carrom or billiard collisions.
Assertion (A): The work done by friction on a body sliding on a rough surface is always negative. Reason (R): Kinetic friction always acts in a direction opposite to the relative sliding of the surfaces in contact.
Answer
A is false but R is true — kinetic friction does indeed oppose relative sliding, so R is correct. But friction is not always negative work: when a block rests on a moving conveyor belt or on an accelerating trolley, friction acts along the direction of the block's displacement and does positive work on it. What is always true is that friction converts mechanical energy into heat for the system as a whole, so the total work done by the pair of friction forces on the two surfaces is negative — but the work on any one body may well be positive. A as stated is therefore too strong.
A pump lifts 500 kg of water through a height of 12 m in 20 s. Calculate its output power (g = 10 m/s²).
Answer
Work done against gravity W = mgh = 500 × 10 × 12 = 60000 J. Power P = W/t = 60000/20 = 3000 W = 3 kW. This is the useful output power; the electrical input power would be higher because of losses in the motor and pipes.
A force F = (3x² + 2) N acts on a particle along the x-axis. Find the work done in moving the particle from x = 1 m to x = 3 m.
Answer
Since the force varies with position, W = ∫F dx from x = 1 to x = 3. W = ∫(3x² + 2)dx = [x³ + 2x] evaluated from 1 to 3. At x = 3: 27 + 6 = 33. At x = 1: 1 + 2 = 3. W = 33 − 3 = 30 J. Graphically this is the area under the force-displacement curve between those limits.
State and prove the work-energy theorem for a variable force. Then use it to find the speed of a 4 kg block, initially at rest, after a net variable force does 72 J of work on it.
Answer
Statement: the net work done by all the forces acting on a body is equal to the change in its kinetic energy, W_net = K_f − K_i = ½mv² − ½mu². Proof for a variable force acting along the x-direction: the work done by force F over an infinitesimal displacement dx is dW = F dx. By Newton's second law F = ma = m(dv/dt). Using the chain rule, m(dv/dt) = m(dv/dx)(dx/dt) = mv(dv/dx). Hence dW = mv(dv/dx)dx = mv dv. Integrating from the initial state (velocity u at position x_i) to the final state (velocity v at position x_f): W = ∫mv dv from u to v = m[v²/2] from u to v = ½mv² − ½mu². That is, W_net = K_f − K_i, which is the work-energy theorem. Note that the derivation nowhere assumes F is constant, so it holds for any force, conservative or not. Application: m = 4 kg, u = 0, W_net = 72 J. 72 = ½(4)v² − 0 = 2v², so v² = 36 and v = 6 m/s.
A 0.5 kg ball is released from rest at the top of a smooth curved track of height 5 m, and at the bottom it strikes a stationary 1.5 kg block, sticking to it. Taking g = 10 m/s², find (a) the ball's speed at the bottom, (b) the common speed after impact, (c) the kinetic energy lost in the collision, and (d) explain where that energy goes.
Answer
(a) On the smooth track only gravity does work, so mechanical energy is conserved: ½mv² = mgh, giving v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s. (b) The collision is perfectly inelastic, so momentum is conserved: m₁v = (m₁ + m₂)V (0.5)(10) = (0.5 + 1.5)V 5 = 2V, so V = 2.5 m/s. (c) Kinetic energy before = ½(0.5)(10)² = 25 J. Kinetic energy after = ½(2.0)(2.5)² = ½ × 2 × 6.25 = 6.25 J. Energy lost = 25 − 6.25 = 18.75 J, which is 75% of the original kinetic energy. (d) The lost kinetic energy is not destroyed: it is converted into other forms during the very short contact — heat in the deformed material of the ball and block, sound energy in the audible thud, and permanent work of deformation. Total energy is still conserved; only mechanical kinetic energy is not, which is exactly what makes the collision inelastic. Momentum survives because the internal deformation forces act in equal and opposite pairs and cancel for the system as a whole.
Read the following and answer the questions that follow: A housing society installs a rooftop water pump that must raise 3000 litres of water per hour to a tank 15 m above the pump. The pump motor is rated 2 kW and is measured to be 60% efficient. Take the density of water as 1000 kg/m³ and g = 10 m/s². (a) Calculate the useful work the pump must do per hour. (b) Find the useful output power required. (c) Determine whether the 2 kW motor is adequate, given its efficiency. (d) Suggest one physical reason why the efficiency is below 100%, and state what happens to the missing energy.
Answer
(a) Mass of water per hour: 3000 litres = 3 m³, so m = ρV = 1000 × 3 = 3000 kg. Useful work per hour W = mgh = 3000 × 10 × 15 = 450000 J = 4.5 × 10⁵ J. (b) Useful output power P_out = W/t = 450000/3600 = 125 W. (c) Efficiency η = P_out/P_in, so the input power actually needed is P_in = 125/0.60 ≈ 208 W. This is far below the motor's 2 kW rating, so the motor is more than adequate — it is in fact heavily oversized for this duty, using only about 10% of its rated capacity. (d) Some input energy is dissipated as heat in the motor windings due to electrical resistance, as heat in the bearings due to friction, and as work against viscous drag and turbulence in the pipes; a little also leaves as sound and vibration. That energy is not lost from the universe — it is degraded into internal energy of the motor, water and surroundings, raising their temperature slightly, which is why total energy is still conserved even though useful mechanical output is less than input.
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