(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

Starting from rest,the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the horizontal is twice the time taken to travel on a smooth plane of the same inclination and same distance. Then the coefficient of kinetic friction is

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$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

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$A$ box is lying on an inclined plane. What is the coefficient of static friction if the box starts sliding when the angle of inclination is $60^\circ$?

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$A$ uniform rope of length $l$ lies on a table. If the coefficient of friction is $\mu$,then the maximum length $l_1$ of the part of this rope which can overhang from the edge of the table without sliding down is

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The force required to just move a body up an inclined plane is double the force required to just prevent the body from sliding down the plane. If the coefficient of friction is $\mu$,then the inclination $\theta$ of the plane is:

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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

Starting from rest,the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the horizontal is twice the time taken to travel on a smooth plane of the same inclination and same distance. Then the coefficient of kinetic friction is

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

$A$ box is lying on an inclined plane. What is the coefficient of static friction if the box starts sliding when the angle of inclination is $60^\circ$?

$A$ uniform rope of length $l$ lies on a table. If the coefficient of friction is $\mu$,then the maximum length $l_1$ of the part of this rope which can overhang from the edge of the table without sliding down is

The force required to just move a body up an inclined plane is double the force required to just prevent the body from sliding down the plane. If the coefficient of friction is $\mu$,then the inclination $\theta$ of the plane is:

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