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#friction

19 public questions tagged with this topic.

Electric Dipole - Moment, Field and Torque

Which property of electric charge explains why the total charge of an isolated system remains constant even when objects

**Electric dipole** consists of charges +q and -q separated by 2a, dipole moment p = q·2a, vector from negative to positive, unit C·m. In uniform field E, torque τ = p × E, magnitude τ = p E sinθ, tending to align p with E, potential energy U = -p·E = -p E cosθ. The conservation of electric charge states that the total charge in an isolated system remains constant over time. When objects are rubbed together, charge is transferred from one to another (e.g., electrons move), but no new charge is created or destroyed. This ensures the net charge of the system stays the

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Work Heat Distinction and Energy Transfer Modes

Which factor primarily causes irreversibility in a process involving friction?

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. Friction converts mechanical energy into heat (dissipation), increasing the system’s or surroundings’ internal energy irreversibly. This energy cannot be fully recovered as work, violating reversibility. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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PHYSICS

A 7 kg block on a horizontal surface ( μ_k = 0.3 ) is pulled by a 4 kg mass over a pulley. A 14 N force opposes the 7 k

Given: A 7 kg block on a horizontal surface ( μ_k = 0.3 ) is pulled by a 4 kg mass over a pulley. A 14 N force opposes the 7 kg block. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 4 kg : 4g - T = 4a Rightarrow 40 - T = 4a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 7 kg : T - f_k - 14 = 7a . Normal: N = mg = 7 × 10 = 70 N . Friction: f_k = 0.3 × 70 = 21 N . Net force: T - 21 - 14 = 7a Rightarrow T - 35 = 7a . Solve: 40 - T = 4a, T - 35 = 7a . Substitute: 40 - (7a + 35) = 4a Rightarrow 40 - 35 - 7a = 4a Rightarrow 5 = 11a . a = 5/11 approx 0.45 m/s² . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

A 4.6 kg block on a horizontal surface ( μ_k = 0.2 ) is connected to a 6.6 kg mass over a pulley. A 9 N force aids the 4

Given: A 4.6 kg block on a horizontal surface ( μ_k = 0.2 ) is connected to a 6.6 kg mass over a pulley. A 9 N force aids the 4.6 kg block. What is the acceleration? (Take g = 10 m/s² ) Formula: For 6.6 kg : 6.6g - T = 6.6a Rightarrow 66 - T = 6.6a. Substitution & Calculation: The 6.6 kg mass descends, pulling the 4.6 kg block with an aiding force. . For 4.6 kg : T + 9 - f_k = 4.6a . Normal: N = mg = 4.6 × 10 = 46 N . Friction: f_k = 0.2 × 46 = 9.2 N . Net force: T + 9 - 9.2 = 4.6a Rightarrow T - 0.2 = 4.6a . Solve: 66 - T = 6.6a, T - 0.2 = 4.6a . Substitute: 66 - (4.6a + 0.2) = 6.6a Rightarrow 66 - 0.2 - 4.6a = 6.6a Rightarrow 65.8 = 11.2a . a = 65.8/11.2 approx 5.88 m/s² . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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Straight Line Motion - Practice Set

A 6.8kg block on a 30∘ incline (μk\=0.15) is pulled upward by a 8.8kg mass over a pulley. What is the acceleration? (Tak

For 8.8kg: 8.8g−T=8.8a⇒88−T=8.8a. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 8.8 a as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

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Straight Line Motion - Practice Set

A 6kg block on a 30∘ incline (μk\=0.25) is connected to a 3kg mass over a pulley. What is the acceleration of the system

Assume 6kg moves down the incline, pulling 3kg up. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 2.83 m/s2 as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

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Straight Line Motion - Practice Set

A 1000kg car on a banked road (θ\=15∘, μs\=0.3) turns at radius 40m. What is the maximum speed without slipping? (Take g

Max speed vmax=rgμs+tan⁡θ1−μstan⁡θ. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 14 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

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Straight Line Motion - Practice Set

A block of mass 2kg is pulled by a force of 10N on a surface with μk\=0.3. What is its acceleration? (Take g\=10m/s2)

Net force Fnet=F−fk, where fk=μkN. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 2 m/s² as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

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