Class 11 Physics, Chapter 5 (NCERT). 15 questions by concept, 3 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
A 5 kg bob starts from rest and descends through 4 m vertically. Air resistance transfers 50 J to internal energy. With g=10 m/s² and no other energy transfer, what is the final kinetic energy?
(b) 150 J
The gravitational-energy decrease is mgh. Subtract the dissipated energy from that decrease to obtain the kinetic-energy gain.
A block completes one closed track of length 12 m. A kinetic-friction force of constant magnitude 5 N opposes the motion at every point. What work does friction do?
(b) -60 J
Friction is opposite to each small displacement around the whole track. Integrating gives −F times path length; zero net displacement does not cancel the negative contributions.
A person carries a 20 kg load at constant height through 5 m horizontally. Their supporting force is vertical. What work does that supporting force do on the load?
(b) Zero, because the force is perpendicular to the displacement.
Work by a specified force is its dot product with the load displacement. The vertical support and horizontal displacement are perpendicular.
Particles A and B are nonrelativistic. B has 16 times A’s mass. Their kinetic energies are 2 J and 8 J, respectively. Which particle is faster?
(c) A, because its kinetic energy per unit mass is greater.
Compare K/m rather than K alone. B has four times the energy but more than four times the mass, so A has the larger speed.
A block slides 20 m along a horizontal surface. An applied force of 100 N points along motion; kinetic friction of 50 N opposes motion. Weight and normal force do no work. What is the change in kinetic energy?
(c) 1000 J
Add work by all forces: W_net=Fs−fs. The work-energy theorem gives ΔK=W_net.
The graph shows the only force component along a particle’s positive x displacement. What work does it do over the full plotted interval?
(c) 200 J
The area under the straight force segment is a trapezium. Its area is (80+20)×4/2 joules, rather than the endpoint force times the interval.
The particle traverses the entire force-position graph from left to right. What is the net work by the plotted force?
(b) 360 J
Positive and negative force sections contribute signed areas. The net work is 80×6−20×6 joules.
A classical particle has the total energy marked by the dashed line. Which open interval in the potential-energy graph is not allowed?
(a) 2 m < x < 4 m
A physical kinetic energy must satisfy K=E−U≥0. The plateau with potential 8 J exceeds total energy 5 J; its open interval is forbidden.
A spring oscillator with k=2 N/m has total energy k(3 m)²/2. A student places it at x=4 m with negative kinetic energy. Which statement is correct?
(a) The proposed position is forbidden because U exceeds E and K cannot be negative.
The turning-point magnitude is 3 m because E=kA²/2. The proposed magnitude 4 m exceeds it, making E−U negative; the classical particle cannot be there.
The potential energy of an ideal spring is defined as zero when relaxed. Its constant is 100 N/m and its signed deformation is x=-2 m. What is its potential energy?
(c) 200 J
U=kx²/2 with the stated relaxed reference. Squaring the negative deformation makes the stored energy positive.
A bob of mass 1 kg on a light string of length 2 m just completes a vertical circle with zero tension only at the top. With g=10 m/s², what is its minimum kinetic energy at the bottom?
(a) 50 J
At the top, zero tension requires v_top²=gL, so K_top=mgL/2. Add the potential rise 2mgL; the bottom kinetic energy is 5mgL/2.
A motor supplies constant mechanical power while a vehicle moves along the force direction. If its speed becomes 3 times as large, what is final driving force / initial driving force?
(c) 1/3
P=Fv for force along the motion. At fixed power, force is inversely proportional to speed.
A particle has initial x-momentum +4 kg·m/s and final x-momentum −4 kg·m/s after an elastic reflection from a massive wall. External impulse on particle plus wall is negligible. What is the wall’s momentum change?
(a) 8 kg·m/s
The particle’s momentum change is −2 times its initial positive momentum. The wall receives the equal opposite impulse, giving a positive change. Elasticity does not conserve each body’s momentum separately.
A cart of mass m and velocity +9 m/s hits a stationary cart of mass 2m in a head-on elastic collision. What is the incident cart’s final signed velocity?
(b) -3 m/s
For a stationary target, v₁=(m₁−m₂)u/(m₁+m₂). The target is heavier, so the incident cart rebounds; velocity exchange with stopping occurs only for equal masses.
A trolley plus sand has mass 325 kg and speed 27 km/h on a frictionless horizontal track. Sand leaks vertically with no horizontal velocity relative to the trolley until 300 kg remains. What happens to trolley speed?
(b) It stays unchanged because the leaving sand produces no horizontal thrust.
The leaving sand carries the same horizontal speed as the trolley at separation. The remaining subsystem loses that sand’s momentum along with its mass; no horizontal thrust changes trolley speed.
Wrong
The brake force does positive work because it acts on the moving car.
Why it feels right. I associate any applied force with a positive contribution.
Right
Work depends on the component of force along displacement.
Worked example
