Time taken by a block to slide down a frictionless inclined plane moving with constant acceleration

Q. A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a₀. The angle between the inclined plane and ground is θ and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ________.

A. √( 2L / ( g sinθ − a₀ cosθ ) )

B. √( 4L / ( g sin2θ − a₀(1 + cos2θ) ) )

C. √( 2L / ( g sin2θ − a₀(1 + cos2θ) ) )

D. √( 4L / ( g cos²θ − a₀ sinθ cosθ ) )

Correct Answer: √( 4L / ( g sin2θ − a₀(1 + cos2θ) ) )

Solution

We analyze the motion in the non-inertial frame of the inclined plane. Since the plane is accelerating leftward with acceleration a₀, a pseudo force m a₀ acts on the block towards the right.

Resolving forces along the incline, the effective acceleration of the block along the plane becomes:

a = g sinθ − a₀ cosθ

The distance travelled by the block along the incline is:

s = L / cosθ

Using kinematics,

s = ½ a t²

Substituting and simplifying using trigonometric identities:

t = √( 4L / ( g sin2θ − a₀(1 + cos2θ) ) )

Hence, the correct expression for time is given by option B.

Solution

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