Static Friction and Kinetic Friction Question | Static Friction and Kinetic Friction JEE Main Question

Q 1 :

A 2 kg brick begins to slide over a surface which is inclined at an angle of 45° with respect to horizontal axis.

The co-efficient of static friction between their surfaces is [2024]

1
0.5
1.7
$\frac{1}{\sqrt{3}}$

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(1)

Slipping starts when, $\tan θ = μ_{s}$ ( $θ$ is angle of respose)

$μ_{s} = \tan 45^{o}$

$\Rightarrow μ_{s} = 1$

Q 2 :

Given below are two statements:

Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.

Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.

In the light of the above statements, choose the most appropriate answer from the given below: [2024]

Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Both Statement I and Statement Il are correct

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(2)

Co-efficient of friction depends on surface in contact So, it depends on material of object.

Q 3 :

A heavy box of mass 50 kg is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.3, then the force of kinetic friction is [2024]

1.47 N
14.7 N
1470 N
147 N

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(4)

Given $m = 50$ kg, $μ_{k} = 0.3$

$f_{k} = μ_{k} N = μ_{k} m g = 0.3 \times 50 \times 9.8 = 147 N$

Q 4 :

A given object takes n times the time to slide down 45° rough inclined plane as it takes the time to slide down an identical perfectly smooth 45° inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is ______ [2024].

$\sqrt{1 - n^{2}}$
$\sqrt{1 - \frac{1}{n^{2}}}$
$1 - n^{2}$
$1 - \frac{1}{n^{2}}$

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(4)

Case-1: No friction

$a = g \sin θ$

$ℓ = \frac{1}{2} (g \sin θ) t_{1}^{2} \Rightarrow t_{1} = \sqrt{\frac{2 ℓ}{g \sin θ}}$

Case-2: With friction

$a = g \sin θ - μ g \cos θ$

$ℓ = \frac{1}{2} (g \sin θ - μ g \cos θ) t_{2}^{2} \Rightarrow t_{2} = \sqrt{\frac{2 ℓ}{(g \sin θ - μ g \cos θ)}}$

$t_{2} = n t_{1} \Rightarrow \sqrt{\frac{2 ℓ}{(g \sin θ - μ g \cos θ)}} = n \sqrt{\frac{2 ℓ}{g \sin θ}}$

$\Rightarrow (\sin θ - μ \cos θ) = \frac{1}{n^{2}} \sin θ$

$\Rightarrow (1 - μ) = \frac{1}{n^{2}} \Rightarrow μ = 1 - \frac{1}{n^{2}}$

Q 5 :

A block of mass m is placed on a surface having a vertical cross-section given by $y = x^{2} / 4$ . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is [2024].

1/4 m
1/2 m
1/6 m
1/3 m

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(1)

$f_{s} = m g \sin θ$ ...(1)

$N = m g \cos θ$ ...(2)

Now, $f_{s} \leq μ N$

$m g \sin θ \leq μ m g \cos θ$

$\tan θ \leq μ$ ...(3)

According to question, $y = \frac{x^{2}}{4}$

$\tan θ = \frac{d y}{d x} = \frac{1}{4} \cdot 2 x = \frac{x}{2}$

Put in (3) $\frac{x}{2} \leq μ \Rightarrow x \leq 2 (0.5)$

$\Rightarrow x \leq 1$ $So y \leq \frac{1^{2}}{4} \Rightarrow y_{\max} = \frac{1}{4} = 0.25$

Q 6 :

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure.

If ${\vec{F}}_{1}$ is the force required to just move the block up the inclined plane and ${\vec{F}}_{2}$ is the force required to just prevent the block from sliding down, then the value of $| {\vec{F}}_{1} |$ $-$ $| {\vec{F}}_{2} |$ is [Use g = 10 $m / s^{2}$ ] [2024]

$25 \sqrt{3}$ N
$5 \sqrt{3}$ N
$\frac{5 \sqrt{3}}{2}$ N
10 N

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(2)

$f_{K} = μ m g \cos θ$

$= 0.1 \times \frac{50 \times \sqrt{3}}{2} = 2.5 \sqrt{3} N$

$F_{1} = m g \sin θ + f_{K} = 25 + 2.5 \sqrt{3}$

$F_{2} = m g \sin θ - f_{K} = 25 - 2.5 \sqrt{3}$

$∴$ $F_{1} - F_{2} = 5 \sqrt{3} N$

Q 7 :

Consider a block and trolley system as shown in the figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in ${ms}^{- 2}$ is ______.

(Consider that the string is massless and unstretchable and the pulley is also massless and frictionless.) [2024]

3
4
2
1.2

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(3)

$f_{k} = μ N = 0.04 \times 20 g = 8 Newton$

$Acceleration a = \frac{60 - 8}{26} = 2 {m/s}^{2}$

Q 8 :

A body of mass 10 kg is moving with an initial speed of 20 m/s. The body stops after 5 s due to friction between the body and the floor. The value of the coefficient of friction is
(Take acceleration due to gravity g = 10 ${ms}^{- 2}$ ) [2023]

0.2
0.3
0.5
0.4

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(4)

$a = - μ g$

$∵$ $v = u + a t$

$0 = 20 + (- μ \times 10) \times 5$

$50 μ = 20$

$μ = \frac{2}{5} = 0.4$

Q 9 :

A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30 N is applied in the direction parallel to the surface of the table, the block slides through a distance of 50 m in an interval of time 10 s. Coefficient of kinetic friction is (given g = 10 ${ms}^{- 2}$ ) [2023]

0.60
0.75
0.50
0.25

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(3)

$S = u t + \frac{1}{2} a t^{2}$

$50 = 0 + \frac{1}{2} \times a \times 100$

$\Rightarrow a = 1 {m/s}^{2}$

$F - μ m g = m a$

$30 - μ \times 50 = 5 \times 1$

$\Rightarrow 50 μ = 25$

$\Rightarrow μ = \frac{1}{2}$

Topic Question Set

Q 1 :

A 2 kg brick begins to slide over a surface which is inclined at an angle of 45° with respect to horizontal axis.

The co-efficient of static friction between their surfaces is [2024]

Q 3 :

A heavy box of mass 50 kg is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.3, then the force of kinetic friction is [2024]

Q 4 :

A given object takes n times the time to slide down 45° rough inclined plane as it takes the time to slide down an identical perfectly smooth 45° inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is ______ [2024].

Q 5 :

A block of mass m is placed on a surface having a vertical cross-section given by $y = x^{2} / 4$ . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is [2024].

Q 8 :

A body of mass 10 kg is moving with an initial speed of 20 m/s. The body stops after 5 s due to friction between the body and the floor. The value of the coefficient of friction is
(Take acceleration due to gravity g = 10 ${ms}^{- 2}$ ) [2023]

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