A block of mass 5 kg is placed on a frictionless incline of angle $30^\circ$. What is the acceleration of the block down the incline? (Take $g = 9.8\, m/s^2$)

Question

A particle of mass $m$ is projected with a velocity $u$ making an angle of $30^\circ$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is _____.

Answer 1

The acceleration of the block on a frictionless incline is determined by the component of gravity acting parallel to the incline. This is calculated using the formula $a = g \sin \theta$, where $a$ is acceleration, $g$ is gravitational acceleration ($9.8\, m/s^2$), and $\theta$ is the incline angle. With $\theta = 30^\circ$, the calculation is:

$a = 9.8 \times \sin 30^\circ$

As $\sin 30^\circ = 0.5$, the acceleration is:

$a = 9.8 \times 0.5 = 4.9\, m/s^2$

The block accelerates down the incline at $4.9\, m/s^2$.

A block of mass 5 kg is placed on a frictionless incline of angle $30^\circ$. What is the acceleration of the block down the incline? (Take $g = 9.8\, m/s^2$)

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The Correct Option is B

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