(C) The component of acceleration due to gravity along the slope is $a_g = g \sin \theta = 10 \sin 30^{\circ} = 5\, m/s^2$.The acceleration due to fri

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(C) The component of acceleration due to gravity along the slope is $a_g = g \sin \theta = 10 \sin 30^{\circ} = 5\, m/s^2$.The acceleration due to friction opposing the motion is $a_f = \mu g \cos \theta = 0.5 \times 10 \times \cos 30^{\circ} = 5 \times 0.866 = 4.33\, m/s^2$.The net acceleration of the car down the slope is $a = a_g - a_f = 5 - 4.33 = 0.67\, m/s^2$.Using the equation of motion $v^2 = u^2 + 2as$,where $u = 6\, m/s$,$a = 0.67\, m/s^2$,and $s = 15\, m$:$v^2 = (6)^2 + 2 \times 0.67 \times 15$$v^2 = 36 + 20.1 = 56.1$$v = \sqrt{56.1} \approx 7.49\, m/s$.

Class 11 Physics Newton's Laws of Motion and Friction Motion (or rest) on Rough Inclined Surface

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$A$ 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 ........ $m/s$. $(g = 10\, m/s^2)$

A
$4$
B
$3$
C
$7.49$
D
$8.45$

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Newton's Laws of Motion and Friction

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Motion (or rest) on Rough Inclined Surface

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F)$ required to move the block up the inclined plane will be

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$A$ block rests on a fixed wedge inclined at an angle $\theta$. The coefficient of friction between the block and the plane is $\mu$. The maximum value of $\theta$ for the block to remain motionless on the wedge is

MediumTS EAMCET 2020

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The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction,when the angle of inclination of the plane is $60^{\circ}$ is

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$A$ block of mass $10 \,kg$ is held at rest against a rough vertical wall $[\mu=0.5]$ under the action of a force $F$ as shown in the figure. The minimum value of $F$ required for it is ............ $N$ $(g=10 \,m/s^2)$.

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$A$ 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 ........ $m/s$. $(g = 10\, m/s^2)$

Similar Questions

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F)$ required to move the block up the inclined plane will be

An insect is crawling in a hemispherical bowl of radius $R$. If the coefficient of friction between the insect and the bowl is $\mu$,then the maximum height to which the insect can crawl in the bowl is

$A$ block rests on a fixed wedge inclined at an angle $\theta$. The coefficient of friction between the block and the plane is $\mu$. The maximum value of $\theta$ for the block to remain motionless on the wedge is

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction,when the angle of inclination of the plane is $60^{\circ}$ is

$A$ block of mass $10 \,kg$ is held at rest against a rough vertical wall $[\mu=0.5]$ under the action of a force $F$ as shown in the figure. The minimum value of $F$ required for it is ............ $N$ $(g=10 \,m/s^2)$.

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