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Standard XII

Physics

Work Energy Theorem Application

A block of ma...

Question

A block of mass 3 $k g$ is released from the rest at top of the inclined plane. If length and height of the inclined plane is 6 m and 4 m respectively. Calculate its speed at the bottom of inclined plane. Assume plane is frictionless.

5 m/s

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6 m/s

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8 m/s

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9 m/s

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10 m/s

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Solution

The correct option is C 9 m/s
Acceleration of the block along the inclined surface= $a = g s i n θ$
$= 10 m / s^{2} \times \frac{4}{6}$
$= \frac{20}{3} m / s^{2}$
Distance traveled along the surface= $s = 6 m$
Thus speed attained by the block= $\sqrt{2 a s} \approx 9 m / s$

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A block of mass 3 kg is released from the rest at top of the inclined plane. If length and height of the inclined plane is 6 m and 4 m respectively. Calculate its speed at the bottom of inclined plane. Assume plane is frictionless.

The correct option is C 9 m/sAcceleration of the block along the inclined surface=a=gsinθ=10m/s2×46=203m/s2Distance traveled along the surface=s=6mThus speed attained by the block=√2as≈9m/s

A block of mass 3 $k g$ is released from the rest at top of the inclined plane. If length and height of the inclined plane is 6 m and 4 m respectively. Calculate its speed at the bottom of inclined plane. Assume plane is frictionless.

The correct option is C 9 m/s
Acceleration of the block along the inclined surface= $a = g s i n θ$
$= 10 m / s^{2} \times \frac{4}{6}$
$= \frac{20}{3} m / s^{2}$
Distance traveled along the surface= $s = 6 m$
Thus speed attained by the block= $\sqrt{2 a s} \approx 9 m / s$