(B) The normal force $N$ acting on the block is given by $N = mg \cos \theta$.Given: $m = 1 \ kg$,$g = 10 \ m/s^2$,$\theta = 60^{\circ}$,and $\mu_s =

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(B) The normal force $N$ acting on the block is given by $N = mg \cos \theta$.Given: $m = 1 \ kg$,$g = 10 \ m/s^2$,$\theta = 60^{\circ}$,and $\mu_s = 0.1$.$N = 1 \times 10 \times \cos(60^{\circ}) = 10 \times 0.5 = 5 \ N$.The frictional force $f$ is given by $f = \mu_s N$.$f = 0.1 \times 5 = 0.5 \ N$.The work done against the frictional force over a distance $d = 10 \ m$ is:$W = f \times d = 0.5 \times 10 = 5 \ J$.

Class 11 Physics Laws of Motion Motion (or rest) on Rough Inclined Surface

Difficult

$A$ block of mass $1 \ kg$ is pushed up a surface inclined to the horizontal at an angle of $60^{\circ}$ by a force of $10 \ N$ parallel to the inclined surface as shown in the figure. When the block is pushed up by $10 \ m$ along the inclined surface,the work done against the frictional force is: $\left[g=10 \ m/s^2, \mu_s = 0.1\right]$

JEE MAIN 2024 All JEE MAIN

A
$5 \sqrt{3} \ J$
B
$5 \ J$
C
$5 \times 10^3 \ J$
D
$10 \ J$

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Class 11

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Chapter

Laws of Motion

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Motion (or rest) on Rough Inclined Surface

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$A$ block of mass $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.

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The upper portion of an inclined plane of inclination $\alpha$ is smooth and the lower portion is rough. $A$ particle slides down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to rough length is $m : n$,the coefficient of friction is

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$A$ homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is:

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Define the condition for a vehicle to be parked on a banked slope without slipping.

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$A$ uniform chain of length $L$ which hangs partially from a table,is kept in equilibrium by friction. The maximum length that can hang without slipping is $l$. Then,the coefficient of friction between the table and the chain is:

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$A$ block of mass $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.

The upper portion of an inclined plane of inclination $\alpha$ is smooth and the lower portion is rough. $A$ particle slides down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to rough length is $m : n$,the coefficient of friction is

$A$ homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is:

Define the condition for a vehicle to be parked on a banked slope without slipping.

$A$ uniform chain of length $L$ which hangs partially from a table,is kept in equilibrium by friction. The maximum length that can hang without slipping is $l$. Then,the coefficient of friction between the table and the chain is:

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