(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 Newton's Laws of Motion and Friction Motion (or rest) on Rough Inclined Surface

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$A$ 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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Newton's Laws of Motion and Friction

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Motion (or rest) on Rough Inclined Surface

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$A$ block is placed on a parabolic shape ramp given by the equation $y = \frac{x^2}{20}$. If the coefficient of static friction $\mu_s$ is $0.5$, then what is the maximum height above the ground at which the block can be placed without slipping (in $m$)?

MediumTS EAMCET 2022

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Difficult

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$A$ block slides down on an inclined surface of inclination $30^{\circ}$ with the horizontal. Starting from rest,it covers $8 \ m$ in the first two seconds. Find the coefficient of friction. (Take $g = 10 \ m/s^2$)

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The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

DifficultAP EAMCET 2013

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$A$ 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:

DifficultJEE MAIN 2024

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$A$ block is placed on a parabolic shape ramp given by the equation $y = \frac{x^2}{20}$. If the coefficient of static friction $\mu_s$ is $0.5$, then what is the maximum height above the ground at which the block can be placed without slipping (in $m$)?

$A$ car is going at a speed of $6\, m/s$ when it encounters a $15\, m$ slope of angle $30^o$. The friction coefficient between the road and tyre is $0.5$. The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is ........ $m/s$. $(g = 10\, m/s^2)$

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