(C) The acceleration $a$ of a block sliding down an inclined plane with friction is given by the formula: $a = g(\sin \theta - \mu \cos \theta)$.Given

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(C) The acceleration $a$ of a block sliding down an inclined plane with friction is given by the formula: $a = g(\sin \theta - \mu \cos \theta)$.Given values are: $g = 10\,m/s^2$,$\theta = 60^\circ$,and $\mu = 0.25$.Substituting these values into the formula:$a = 10 \times (\sin 60^\circ - 0.25 \times \cos 60^\circ)$$a = 10 \times (\frac{\sqrt{3}}{2} - 0.25 \times \frac{1}{2})$$a = 10 \times (0.866 - 0.125)$$a = 10 \times 0.741$$a \approx 7.4\,m/s^2$.

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

Medium

$A$ block is lying on an inclined plane which makes $60^\circ$ with the horizontal. If the coefficient of friction between the block and the plane is $0.25$ and $g = 10\,m/s^2$,then the acceleration of the block when it moves along the plane will be ........ $m/s^2$.

A
$2.50$
B
$5.00$
C
$7.4$
D
$8.66$

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Class 11 Questions

Class 11

Physics Questions

Chapter

Newton's Laws of Motion and Friction

Subtopic

Motion (or rest) on Rough Inclined Surface

$A$ block of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

MediumJEE MAIN 2023

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$A$ piece of ice slides down a rough inclined plane at $\theta=45^{\circ}$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of friction between ice and incline?

DifficultAIIMS 2018

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Find the work done by friction if a $1 \, kg$ block reaches the end of an inclined plane of length $10 \, m$ and inclination $30^{\circ}$ with constant velocity.

Medium

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$A$ given object takes $n$ times more time to slide down a $45^o$ rough inclined plane as it takes to slide down a perfectly smooth $45^o$ incline. The coefficient of kinetic friction between the object and the incline is

Difficult

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The upper half of an inclined plane of inclination $\theta$ is perfectly smooth,while the lower half is rough. $A$ body starting from rest at the top comes back to rest at the bottom. The coefficient of friction $\mu$ for the lower half is given by:

DifficultAIEEE 2005

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$A$ block is lying on an inclined plane which makes $60^\circ$ with the horizontal. If the coefficient of friction between the block and the plane is $0.25$ and $g = 10\,m/s^2$,then the acceleration of the block when it moves along the plane will be ........ $m/s^2$.

Similar Questions

$A$ block of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

$A$ piece of ice slides down a rough inclined plane at $\theta=45^{\circ}$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of friction between ice and incline?

Find the work done by friction if a $1 \, kg$ block reaches the end of an inclined plane of length $10 \, m$ and inclination $30^{\circ}$ with constant velocity.

$A$ given object takes $n$ times more time to slide down a $45^o$ rough inclined plane as it takes to slide down a perfectly smooth $45^o$ incline. The coefficient of kinetic friction between the object and the incline is

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth,while the lower half is rough. $A$ body starting from rest at the top comes back to rest at the bottom. The coefficient of friction $\mu$ for the lower half is given by:

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