NEET Friction MCQs & PYQs

Practice NEET Friction Questions

Question 1:

easy

Which one of the following statements is incorrect?

(2018)

Friction force always opposes relative motion. Limiting static friction \( \text{f}_{\text{s,max}} \) is proportional to normal reaction \( \text{N} \), so \( \text{f}_{\text{s,max}} = \mu_{\text{s}}\text{N} \). Rolling friction is generally much smaller than sliding friction. The coefficient of friction \( \mu \) is a ratio of forces (friction force to normal force), thus it is a dimensionless quantity. Therefore, option (d) is incorrect.

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Question 2:

moderate

A conveyor belt is moving at a constant speed of \( 2 \text{ m/s} \). A box is gently dropped on it. The coefficient of friction between them is \( \mu = 0.5 \). The distance that the box will move relative to belt before coming to rest on it, taking \( g = 10 \text{ ms}^{-2} \).

(2011 Mains)

Initial relative speed of box \( u = 2 \text{ m/s} \). Friction force \( f = \mu mg \). Acceleration \( a = \frac{f}{m} = \mu g = 0.5 \times 10 = 5 \text{ m/s}^2 \). Using \( v^2 = u^2 - 2as \) (since friction causes deceleration). With \( v = 0 \), distance \( s = \frac{u^2}{2a} = \frac{(2)^2}{2 \times 5} = \frac{4}{10} = 0.4 \text{ m} \).

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Question 3:

easy

A block of mass \(10\text{ kg}\) placed on rough horizontal surface having coefficient of friction \(\mu = 0.5\), if a horizontal force of \(100text{ N}\) acting on it then acceleration of the block will be:

(2002)

Calculate max friction \(f = \mu mg\). Compare with applied force (F). If (F > f), then (a = (F-f)/m). Here, (f = 0.5 \times 10 \times 10 = 50\text{ N}). So, (a = (100-50)/10 = 5\text{ m/s}^2).

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Question 4:

easy

On the horizontal surface of a truck a block of mass \(1\text{ kg}\) is placed \((\mu = 0.6)\) and truck is moving with acceleration \(5\text{ m/s}^2\) then the frictional force on block will be:

(2001)

Required force for block to accelerate with truck is \(F = ma_{truck}\). Maximum static friction is \(f_{s,max} = \mu mg\). If \(F \leq f_{s,max}\), friction equals (F). Here, \(F = 1 \times 5 = 5\text{ N}\) and \(f_{s,max} = 0.6 \times 1 \times 10 = 6\text{ N}\). Since \(5 \leq 6\), the frictional force is \(5\text{ N}\).

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Question 5:

easy

Consider a car moving along a straight horizontal road with a speed of \(72\text{ km/h}\). If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance, in which the car can be stopped is (taking \(g = 10\text{ m/s}^2)\):

(1992)

Convert speed to m/s: \(v = 72 \times 5/18 = 20\text{ m/s}\). Deceleration due to friction is \(a = \mu_s g = 0.5 \times 10 = 5\text{ m/s}^2\). Using \(v^2 = u^2 + 2as\), \(0 = (20)^2 - 2(5)s\). So, \(10s = 400\) and \(s = 40\text{ m}\).

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Question 6:

moderate

A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is:

(1991)

For equilibrium, the weight of the hanging part must equal the maximum static friction on the table. If (x) is the fraction hanging, then \(xMg = mu(1-x)Mg\). So, \(x = \mu(1-x)\). This gives \(x = \frac{\mu}{1+\mu}\). With \(\mu = 0.25\), \(x = \frac{0.25}{1+0.25} = \frac{0.25}{1.25} = 0.20\) or (20%).

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