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Question

A frictionless inclined plane of length l having inclination &theta; is placed inside a lift which is accelerating downward with an acceleration a($<g$) . If a block is allowed to move, down the inclined plane, from rest, then the time taken by the block to slide from top of the inclined plane to the bottom of the inclined plane is

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
	**Primary Concept**: Motion on an inclined plane with a non-inertial frame of reference
	**Related Topics**: Newton's second law, pseudo force, kinematics of uniformly accelerated motion.
	**Prerequisites**: Understanding of forces, Newton's laws, and equations of motion.
## SOLUTION PATHWAY
**Given Information**:
	Length of inclined plane = l
	Inclination angle = θ
	Downward acceleration of lift = a
	Initial velocity of block = 0
**Method & Steps**:
	1.  **Calculate the effective acceleration:**
		Reasoning: Since the lift is accelerating downwards, a pseudo force acts upwards on the block. The effective acceleration down the incline is the component of (g-a) along the incline.
		Formula/Rule:  Effective acceleration along the incline, $a_{eff} = (g-a)sinθ$
	2.  **Use kinematics equation to find the time:**
		Working: We can use the equation of motion $s = ut + \frac{1}{2}at^2$. Here, initial velocity (u) = 0, distance (s) = l, and acceleration (a) = $a_{eff} = (g-a)sinθ$.
		Key Point: Substitute the values and solve for 't'.
	3.  **Solve for time:**
		Result: $l = 0 + \frac{1}{2}(g-a)sinθt^2$. Solving for t, we get $t = \sqrt{\frac{2l}{(g-a)sinθ}}$
		Verification: The units are consistent.
 ## OPTION ANALYSIS
[A] :x: This option does not account for the lift's acceleration or the incline.
[B] :x: This option accounts for the lift's acceleration but not the incline.
[C] :x: This option does not use correct effective acceleration and neglects the incline.
[D] ✓: This option correctly calculates the effective acceleration and uses the correct formula for time.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Forgetting the pseudo force:** Not considering the pseudo force due to the lift's acceleration leads to an incorrect effective acceleration.
		2. **Incorrect component of acceleration:** Not taking the component of effective acceleration along the incline.
	**Quick Tips**:
		1. **Draw a free body diagram:** This will help visualize the forces and determine the net acceleration.
		2. **Remember the formula:** $s = ut + \frac{1}{2}at^2$ is essential for kinematics problems.

Mr. ANIL KUMAR.G · Answer

$\sqrt {\frac{{2l}}{{\rm{g}}}} $

Mr. ANIL KUMAR.G · Answer

$\sqrt {\frac{{2l}}{{{\rm{g}} - a}}} $

Mr. ANIL KUMAR.G · Answer

$\sqrt {\frac{{2l}}{{{\rm{g}} + a}}} $

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\sqrt {\frac{{2l}}{{\left( {{m{g}} - a} ight)\sin {m{	heta }}}}} $"

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