Maximum height reached by the block (A) (h0)/(2) (B) h0 (C) (3 h0)/(4) (D) (h0)/(4)

Name: Problem 88, Chapter 8: A Complete Resource Book in Physics for JEE Main
Uploaded: 2021-10-07T05:10:48-08:00
Duration: 5 min 16 s
Channel: Mahipal Kumawat
Description: Maximum height reached by the block (A) (h0)/(2) (B) h0 (C) (3 h0)/(4) (D) (h0)/(4)

Maximum height reached by the block (A) (h0)/(2) (B) h0 (C)...

Key Concepts

Gravitational Potential Energy

This refers to the energy an object possesses due to its position in a gravitational field. In vertical motion problems, the gain in gravitational potential energy as the block rises is directly related to the height ascended, making it an essential concept for calculating the maximum height reached by the block.

Kinematics of Vertical Motion

This involves the equations that describe the motion of objects under constant acceleration, such as gravity. When a block is projected upward, its velocity decreases until it reaches zero at the maximum height. The kinematic equations help relate the initial velocity, acceleration due to gravity, and displacement, thereby enabling the calculation of the maximum height.

Conservation of Mechanical Energy

This concept states that in the absence of non-conservative forces like friction, the total mechanical energy (kinetic plus potential) of a system remains constant. It is crucial when analyzing problems where energy transforms from one form to another, such as when a block's kinetic energy is entirely converted into gravitational potential energy at its maximum height.

Transcript

00:01 On in the figure, the liquid density varies as row equal to row 0, 4 minus 3 h by h0, where h is the height of the liquid from the bottom.

00:09 And a solid block has mass m and density d equal to 5 by 2 row 0 is at the bottom of the liquid.

00:19 Now this block starts moves upwards, then in that case the f net on the block, resultant force will be equal to f3.

00:30 Up thrust force minus the weight of the block mg.

00:33 Up thrust force can be written as v row g minus mg.

00:38 V can be written as mass of the block divide density of block row g minus m.

00:45 That can be written as m divide 2 by 5 row 0.

00:51 Multiply this row can be written as row 0, 4 minus 3 h by h0, minus 3 h0, minus m g minus m g this is multiplied by g so if you put the value here then we get uh it becomes two m g 2 mg by 5 2 mg by 5 4 minus 3 h by h 0 minus m g minus m g so this is f net f net now if you put h equal to h 0 by 2 that means at the of the liquid at the mid of the liquid this f net comes out to be zero f net comes out to be zero that means this is the mean position mean position of the motion mean position of the motion now if we move upward if if we move upward from if we move upward upward from the main position from h equal to h0 by 2 by a distance x by a distance x then the f net then the f net will be equal to 2mg by 5 4 minus 3 now the distance will be h 0 by 2 plus x from the bottom divide h 0 minus m g minus m g so if you solve this we get minus 6mg minus 6mg x divide 5 h0 5 h 0 that means here f is proportional to minus x and when this condition occurs then we can say that this motion is simple harmonic motion this motion is simple harmonic motion so in the given options in question number 85 question number 85 the option b is the right answer where the motion is of type of shm now if we compare the f this f equal to minus 6mg x by 5 h0 then we can say that value of omega that will be equal to under root 6g by 5 .0.

03:05 So from here i can write the time period t that will be equal to 2 pi by omega...

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