JEE Main 2022PhysicsWork Power EnergyHardNumerical

JEE Main 2022 — Work Power Energy Question with Solution

JEE Main 2022 (27 Jul Shift 2)

Question

Two inclined planes are placed as shown in figure.
A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t2+1s. The value of t is _____ (use g=10 m s-2)

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

From energy conservation at point A and B, 

12mv02=mgh

⇒v0=gh=10×10

⇒v0=102 m s-1

For A→B 

At B,  v=0

Acceleration of a particle moving on a smooth incline is gsinθ. Therefore, along AB, a=-gsin45°=-102 m s-2.

Using equation of motion,

v=u+at1

0=102-102t1⇒t1=2 s

For B→C

Using second equation of motion, 

s=ut2+12at22

⇒10sin30°=1210sin30°t22

t2=22 s

So total time T=t1+t2 =22+2

=22+1 s

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About this question

This is a previous-year question from JEE Main 2022, covering the Work Power Energy chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.