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a block of mass 2 kg, kept at origin at t=0 is having velocity 4√6 in positive x-direction. Its potential energy is defined as U= -x^3+6x^2+15 . Its velocity when the applied force 8s minimum is.....
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a block of mass 2 kg, kept at origin at t=0 is having velocity 4√6 in ...
Problem:

A block of mass 2 kg, kept at origin at t=0 is having velocity 4√6 in positive x-direction. Its potential energy is defined as U= -x^3 6x^2 15. Its velocity when the applied force 8s minimum is.....


Solution:

Given, mass of the block, m = 2kg, initial velocity, u = 4√6 m/s, and potential energy, U = -x^3 6x^2 15.


Step 1: Calculation of Kinetic Energy

Initial Kinetic Energy of the block, K = (1/2)mv^2 = (1/2) x 2 x (4√6)^2 = 96 J


Step 2: Calculation of Force

Given, potential energy, U = -x^3 6x^2 15.

Force, F = -dU/dx = -(-3x^2 + 12x) = 3x^2 - 12x N


Step 3: Calculation of Velocity

Minimum force is applied when F = 0.

So, 3x^2 - 12x = 0

x(3x - 12) = 0

x = 0 or x = 4


As the block is initially at the origin, x = 0 is the initial position.

Final position, x = 4m

Work done by the force, W = U(0) - U(4) = -15 J


From the work-energy theorem,

W = Kf - Ki

Where Kf is the final kinetic energy and Ki is the initial kinetic energy.

So, Kf = W + Ki = -15 + 96 = 81 J


Final velocity, v = √(2Kf/m) = √(2x81/2) = 9 m/s


Answer:

The velocity of the block when the applied force is minimum is 9 m/s.
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a block of mass 2 kg, kept at origin at t=0 is having velocity 4√6 in ...
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a block of mass 2 kg, kept at origin at t=0 is having velocity 4√6 in positive x-direction. Its potential energy is defined as U= -x^3+6x^2+15 . Its velocity when the applied force 8s minimum is.....
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