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Sticking of Block With Accelerated Cart

NEET > Physics > Laws of Motion > Friction > Sticking of Block With Accelerated Cart

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NEET Physics — Friction

Sticking of Block With Accelerated Cart – Complete Notes, Revision, Important Questions & Downloads

Sticking of Block With Accelerated Cart covers the TOC subtopic Minimum Acceleration for Block to Stick. A small block of mass m is placed against the vertical face of a cart of mass M. The cart accelerates horizontally to the right. In the non-inertial frame, a pseudo-force ma acts on the block to the left, pressing it against the vertical cart wall. Friction (μ × normal force = μma) acts upward, supporting the block's weight mg. For the block not to slide down: μma ≥ mg → a ≥ g/μ. Minimum acceleration: a_min = g/μ. The minimum external force on the system: F_min = (M+m) × g/μ. NEET tests this as a direct substitution problem.

⬇ Download Notes PDFView Important Questions →
Minimum Acceleration for Block to StickFriction Ch.5Pseudo-Force Analysis
Expected QuestionsQ
0–1
NEET asks: (1) find a_min = g/μ for the block to stick to the vertical face, (2) find the minimum force F on the (cart+block) system, (3) conceptual: what happens if a < g/μ (block slides down). Occasionally: compare two blocks of different masses attached to the same cart.
Time Required⏱
20 min
10 min: understand the pseudo-force setup. In the non-inertial frame (cart's frame), pseudo-force = ma (horizontal, into the wall). This is the normal force on the block. Friction = μ × ma (vertical upward). Condition: μma ≥ mg → a ≥ g/μ. 10 min: find minimum force F = (M+m) × a_min = (M+m)g/μ. Practice 3 numericals.
Difficulty⚡
Medium
The key conceptual step is understanding that the cart's acceleration provides the horizontal normal force on the block (through the wall). In the inertial ground frame: the wall pushes the block horizontally with force N=ma (Newton's 3rd law reaction to block pressing on wall). Friction = μN = μma (upward). For equilibrium: μma = mg → a = g/μ.
NRI USA Curriculum GapUS
Medium
AP Physics 1 covers pseudo-forces in accelerating frames but rarely presents the 'block on vertical face of cart' configuration as a named problem. NRI students need to correctly identify that the cart's acceleration creates the normal force on the block's vertical surface — the conceptual leap that friction from a vertical surface can support weight.
1Subtopics
4+Practice Questions
4Free Downloads
20 minPrep Time
⬇ Get Free Downloads

NEET Weightage — Sticking of Block With Accelerated Cart

Friction (Chapter 5)
NEET YearQuestions from this TopicBarMarks
20241
 
1 Q
4
20231
 
1 Q
4
20220
 
0 Q
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20210
 
0 Q
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20201
 
1 Q
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20190
 
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6-Year Total (2019–2024)0–2 0–8
GROUND FRAME ANALYSIS: Cart (mass M) + block (mass m) system. External horizontal force F applied to cart. Both accelerate at a = F/(M+m) (if block doesn't slip). Block (mass m) has net horizontal acceleration a. Only force providing this horizontal acceleration is the normal force N from the cart wall pressing on the block. So N = ma. Friction on block from cart wall: f = μN = μma (directed vertically upward, opposing block's tendency to fall). For block not to slide down: f ≥ mg → μma ≥ mg → a ≥ g/μ.
MINIMUM ACCELERATION: a_min = g/μ. At this minimum: friction exactly equals weight: f = μma_min = μm(g/μ) = mg ✓. If a < g/μ: f = μma < mg — block slides down. If a > g/μ: f = μma > mg — block is held up with excess friction (it is in static friction regime, not the limiting case). MINIMUM FORCE: For the entire (M+m) system to have acceleration a_min = g/μ: F_min = (M+m)×a_min = (M+m)g/μ.

NON-INERTIAL (CART) FRAME: In cart's frame, block is stationary (it's in the cart's frame). Pseudo-force on block = ma acting to the LEFT (opposite to cart's acceleration direction, which is to the right). This pseudo-force presses the block into the RIGHT wall of the cart. Normal force N = ma (from the wall on the block). Friction = μN = μma (upward). Weight = mg (downward). For equilibrium: μma = mg → a = g/μ. Both frames give the same result (as expected).
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12
Total Marks (6 yrs)
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How to Solve the Block-on-Vertical-Cart-Face Problem

1

Step 1 — Identify the normal force The normal force on the block from the cart's vertical face = ma (the block must accelerate horizontally with the cart; only the wall can provide this force). Write N = ma.

2

Step 2 — Write the friction condition Friction supports the block: f = μN = μma (upward). Weight pulls block down: mg. For block not to slide: μma ≥ mg → a ≥ g/μ.

3

Step 3 — Find minimum acceleration and force a_min = g/μ. Minimum force on system: F_min = (M+m) × a_min = (M+m)g/μ.

Study Materials — Sticking of Block With Accelerated Cart

PDF · Cheat Sheet · MCQ Set · PYQ
📘
Full Notes
Block on vertical face of accelerating cart. N=ma (normal force from wall). Friction=μma (upward). Condition: μma≥mg → a≥g/μ. a_min=g/μ. F_min=(M+m)g/μ. Independent of block mass m (a_min=g/μ has no m).
1 subtopicPseudo-force / ground frameDirect formula
Download Notes
📗
Formula Sheet
N=ma; f=μma; a_min=g/μ; F_min=(M+m)g/μ. Key: a_min independent of block mass m.
4 formulas1 pageQuick reference
Download Sheet
📙
MCQ Practice
8 MCQs: compute a_min, compute F_min, compare different μ or different block masses, conceptual frame-switching questions.
8 MCQs2 formulasSolved
Download MCQs
📒
PYQ
NEET previous year questions on block sticking to accelerating cart with complete solutions.
3+ year-tagged Qs2015–2024Solved
Download PYQs

Subtopics — Sticking of Block With Accelerated Cart

2-Column Table
Column AColumn B
Minimum Acceleration for Block to Stick↗

Rapid Revision — Sticking of Block With Accelerated Cart

Concept → Trap → Example

1) Minimum Acceleration for Block to Stick — Complete Analysis

Minimum Acceleration for Block to Stick

SETUP: A cart of mass M is on a smooth horizontal surface. A small block of mass m is placed against the vertical (left or right) face of the cart — the block touches the cart's wall, not the floor. The cart is pushed horizontally (to the right) by an external force F. The block and cart accelerate together at a = F/(M+m) (assuming they move as a unit). THE QUESTION: What is the minimum acceleration a such that the block does NOT slide down the cart's face? GROUND FRAME ANALYSIS: For the block (mass m) to accelerate horizontally at the same rate a as the cart, the cart's wall must push the block horizontally with a net horizontal force = ma. This is the NORMAL FORCE from the wall on the block: N = ma. Friction: between block and cart wall — since the block tends to slide downward (weight mg acts down), the kinetic/static friction acts UPWARD. Friction force f = μN = μma (upward, from cart wall on block). CONDITION FOR BLOCK TO STAY: Upward friction ≥ downward weight: f ≥ mg → μma ≥ mg → a ≥ g/μ. MINIMUM ACCELERATION: a_min = g/μ. AT THE MINIMUM: f = μm × (g/μ) = mg (friction exactly balances weight). Excess acceleration: if a > g/μ, friction > mg, block is held more firmly — static friction, not kinetic. MINIMUM FORCE: F_min produces a_min for the whole system: F_min = (M+m) × a_min = (M+m)g/μ.

  • IMPORTANT: a_min = g/μ is independent of the block's mass m. Whether the block is a 1 kg block or a 100 kg block, the same minimum acceleration is required to hold it against the cart's face. This is analogous to stopping distance being independent of mass — the friction force (μma) and the weight (mg) both scale with m, so m cancels in the condition μma ≥ mg → a ≥ g/μ.
  • WHAT IF BLOCK HAS MASS M (same as cart)? The block mass doesn't affect a_min = g/μ. However, the minimum FORCE does depend on total mass: F_min = (M+m)g/μ. If block mass m increases, F_min increases (need more force to accelerate a heavier system to the same a_min). Common confusion: 'bigger block needs more force' is TRUE for the force on the system, but NOT for the minimum acceleration.
  • NON-INERTIAL FRAME SHORTCUT: In the cart's (non-inertial) frame, the block appears stationary. Three forces: weight mg (down), pseudo-force ma (horizontal, away from the direction of cart's acceleration — into the wall), and friction f (up). Equilibrium: horizontal: N = ma (pseudo-force balanced by normal); vertical: f = mg. At limit: f = μN = μma = mg → a = g/μ. This non-inertial frame analysis gives the same result and is sometimes faster to set up.
Example (NEET-style)Cart mass M = 4 kg, block mass m = 1 kg, μ = 0.25 between block and cart. a_min = g/μ = 10/0.25 = 40 m/s². F_min = (M+m) × a_min = (4+1)×40 = 200 N. If another block of mass m' = 2 kg replaces the 1 kg block, a_min is still 40 m/s² but F_min = (4+2)×40 = 240 N. The minimum acceleration is unchanged; only the required force changes because the total system mass changes.

US Curriculum Gaps — Sticking of Block With Accelerated Cart

Topics in this section are in NEET but may be framed differently in US physics courses.

Pseudo-Force Analysis in Non-Inertial Frames in AP Physics 1

AP Physics 1 covers Newton's laws in inertial frames and introduces the concept of fictitious forces qualitatively. The specific problem of a block on a vertical cart face — where the cart's acceleration provides the horizontal normal force — is not a standard AP Physics 1 problem type. NRI students need to explicitly practise identifying that the normal force on the block's vertical surface equals ma (not mg), as this is counterintuitive compared to floor-block problems where N = mg.

  • AP Physics 1: pseudo-forces introduced qualitatively; block-on-vertical-cart-face not a standard problem
  • NEET: a_min=g/μ is a direct-formula question — know the derivation via N=ma, f=μma, μma=mg
  • Key insight: normal force on block = ma (horizontal), not related to gravity — vertical friction supports weight

Force Analysis on Objects in Accelerating Reference Frames in University Physics

Halliday & Resnick covers pseudo-forces and accelerating reference frames in the dynamics chapters. The block-on-vertical-cart configuration is sometimes presented as a worked example. However, NEET versions often include the minimum total force F = (M+m)g/μ calculation and comparison of two blocks of different masses (showing mass-independence of a_min), which are extensions beyond typical university textbook treatments.

  • US university: pseudo-force in accelerating frames covered; vertical-cart-face occasionally as example
  • NEET: compute a_min=g/μ and F_min=(M+m)g/μ in 30 seconds — practise until automatic
  • Mass-independence of a_min is a conceptual test — two blocks of different masses need the same a_min

NEET-Style Practice — Sticking of Block With Accelerated Cart

4 Questions
1When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left. This presses the block against the cart wall. If μ = 0.4, what is the minimum acceleration for the block not to fall?Minimum Acceleration for Block to Stick
20 m/s²
25 m/s²
4 m/s²
40 m/s²
a_min = g/μ = 10/0.4 = 25 m/s². Answer: Option B. Derivation: N = ma (normal force from wall). Friction = μN = μma (upward). Condition: μma ≥ mg → a ≥ g/μ = 10/0.4 = 25 m/s². Option A (20 m/s²): μ×20 = 0.4×20 = 8 m/s² < g — friction only provides 8 N per kg, insufficient to hold the block (needs 10 N/kg). Option D (40 m/s²): μ/g = 0.4/10 = 0.04 — this confuses μ/g with g/μ.
2A cart (M = 3 kg) pushes a block (m = 1 kg) against its vertical face. The coefficient of friction between block and cart is μ = 0.5. What is the minimum force the cart must exert on the system?Minimum Force
40 N
80 N
20 N
160 N
a_min = g/μ = 10/0.5 = 20 m/s². F_min = (M+m)×a_min = (3+1)×20 = 80 N. Answer: Option B. Option A (40 N) gives a = 40/(3+1) = 10 m/s². Check: μ×N = μ×m×a = 0.5×1×10 = 5 N < mg = 10 N — insufficient. The system needs F = 80 N to produce a = 20 m/s². At this acceleration: N = ma = 1×20 = 20 N. Friction = 0.5×20 = 10 N = mg ✓.
3This is the minimum acceleration of the cart so that the block does not fall. If the μ between the block and cart wall doubles (from μ to 2μ), the new minimum acceleration is:Effect of Doubling μ
a_min/2
2 × a_min
4 × a_min
a_min (unchanged)
a_min = g/μ. If μ doubles to 2μ: new a_min = g/(2μ) = a_min/2. Answer: Option A. Higher friction means less acceleration needed to hold the block — the block can be held even with a smaller normal force. This makes intuitive sense: rougher cart wall = friction support available at lower N = lower acceleration.
4Two blocks of mass m and 2m are placed against the same vertical face of a cart (one after another, separately). What is the ratio of their minimum accelerations?Mass Independence
1:2
2:1
1:1 (both need the same a_min)
Cannot be determined
For both blocks: a_min = g/μ, which is independent of mass. The block of mass m needs a_min = g/μ, and the block of mass 2m also needs a_min = g/μ. Ratio = 1:1. Answer: Option C. The minimum acceleration is mass-independent because both the required friction (mg) and the normal force (ma) are proportional to mass — when you write μma ≥ mg, the m cancels, leaving a ≥ g/μ.

Practice Problems — Sticking of Block With Accelerated Cart

Click "Reveal Answer" after attempting
1A block of mass 2 kg is placed on the vertical face of a cart (M = 8 kg) moving horizontally. μ = 0.2 between block and cart. Find: (a) a_min, (b) F_min on the cart+block system, (c) Normal force on block at a_min.
(a) 50 m/s²; (b) 500 N; (c) N = 100 N
(a) 50 m/s²; (b) 250 N; (c) N = 50 N
(a) 2 m/s²; (b) 20 N; (c) N = 4 N
(a) 50 m/s²; (b) 1000 N; (c) N = 200 N
👁 Reveal Answer
a_min=g/μ=10/0.2=50 m/s². F_min=(M+m)×a_min=(8+2)×50=500 N. N=m×a_min=2×50=100 N. Friction=μN=0.2×100=20 N=mg=2×10=20 N ✓. Answer: Option A.
2At a_min = g/μ, the block is on the verge of sliding. If the cart's acceleration is increased to 2a_min, what is the static friction force on the block?
2mg (static friction doubles)
mg (static friction equals weight, friction is not at limiting value)
2μmg (kinetic friction at higher speed)
μm(2a_min) = 2μmg (always at limiting value)
👁 Reveal Answer
At a = 2a_min = 2g/μ: N = ma = m×2g/μ = 2mg/μ. Maximum possible friction = μN = μ×2mg/μ = 2mg. But block's weight = mg. Since 2mg > mg, the block doesn't need maximum friction — static friction self-adjusts to exactly mg (just enough to support the weight). Friction = mg, not 2mg. Answer: Option B. Friction is at its limiting value (μN) only at exactly a_min. Above a_min, friction is less than its maximum — it equals mg and uses only part of the available friction.
3A 500 g block is pressed against the vertical wall of a cart by a horizontal push. μ = 0.5. What is the minimum force needed if the cart has mass 1.5 kg? (g = 10 m/s²)
20 N
40 N
30 N
10 N
👁 Reveal Answer
m=0.5 kg, M=1.5 kg, μ=0.5. a_min=g/μ=10/0.5=20 m/s². F_min=(M+m)×a_min=(1.5+0.5)×20=2×20=40 N. Answer: Option B. At this force: a=20 m/s². N=ma=0.5×20=10 N. Friction=μN=0.5×10=5 N=mg=0.5×10=5 N ✓.
4For the block-on-vertical-cart problem, if the coefficient of friction changes from μ to μ/3, by what factor does the minimum force on the entire system change? (Cart mass M, block mass m.)
F increases by factor 1/3
F increases by factor 3
F is unchanged
F increases by factor 9
👁 Reveal Answer
Original F_min = (M+m)g/μ. New μ' = μ/3: new F_min = (M+m)g/(μ/3) = 3(M+m)g/μ = 3×original F_min. F increases by factor 3. Answer: Option B. Physical reason: tripling the roughness would allow the block to stick at 1/3 the acceleration. But reducing roughness to 1/3 requires tripling the acceleration (and hence force) to maintain the block's grip on the wall.

Physics — Friction Revision Checklist

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FAQ — Sticking of Block With Accelerated Cart

Notes · Downloads · Revision · Important Questions
What is the minimum acceleration for a block to stick to the vertical face of a cart?
a_min = g/μ, where g is gravitational acceleration and μ is the coefficient of static friction between the block and the cart's vertical face. At this acceleration, the friction force (= μ × N = μ × ma) exactly equals the block's weight mg.
Why does the normal force on the block equal ma?
The block is against the cart's vertical (not horizontal) face. The only horizontal force on the block is the normal force N from the cart wall. For the block to accelerate horizontally at the same rate a as the cart, Newton's 2nd law gives N = ma. This normal force is NOT related to gravity — it comes from the cart's acceleration.
How does the minimum force F_min relate to a_min?
F_min = (M+m) × a_min = (M+m)g/μ. The external force must accelerate the entire system (cart + block) at a_min = g/μ. The total system mass is (M+m), so F = (M+m)a. At a_min: F_min = (M+m)g/μ.
Is a_min affected by the block's mass?
No. a_min = g/μ is independent of the block's mass. When you write the condition μma ≥ mg, the mass m cancels. Both the required friction (mg) and the available friction (μma) are proportional to m, so m doesn't affect the minimum acceleration. However, the minimum FORCE F_min = (M+m)g/μ does depend on the block's mass m.
What happens if the cart decelerates (slows down) instead of accelerating?
If the cart decelerates (accelerates in the direction opposite to motion), the pseudo-force on the block would push it away from the rear wall (or toward the front wall). The analysis depends on which wall the block is against. If the block is against the front wall and the cart decelerates, the block presses against the front wall — the same analysis applies with the same formula a_min = g/μ, but using the magnitude of deceleration.
What if the cart moves on a rough surface (not frictionless)?
If the cart-surface friction is present, the problem becomes more complex. The applied force F must overcome both the cart-surface friction AND provide the acceleration. For the block sticking condition, only the acceleration a of the system matters (a_min = g/μ between block and cart). The floor friction affects what force F is needed but not the a_min condition itself.
If a = 2 × a_min, what is the friction force on the block?
Normal force N = m × 2a_min = 2mg/μ. Maximum available friction = μN = 2mg. Block's weight = mg. Since maximum friction (2mg) > weight (mg), the block doesn't need maximum friction. Static friction self-adjusts to exactly mg (just enough to support the weight). Friction = mg, not 2mg. The block is well within the static friction regime at a = 2a_min.
Can I solve this using energy methods?
For finding a_min, energy methods don't directly apply (we need force balance, not energy balance). However, for finding the work done by friction or by the applied force in moving a certain distance, energy methods work. The minimum acceleration condition requires force equilibrium (vertical: friction = weight), which is a force-balance — not resolvable via energy.
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