A block of mass 2 kg is kept at origin at t=0 and is having velocity 45√ m/s in positive x-direction. The only force acting on it is a conservative and its potential energy is defind as U=– x3 + 6x2 + 15 (SI units). Its velocity when its acceleration is minimum after t=0 is-
Question
A block of mass 2 kg is kept at origin at t=0 and is having velocity 45√ m/s in positive x-direction. The only force acting on it is a conservative and its potential energy is defind as U=– x3 + 6x2 + 15 (SI units). Its velocity when its acceleration is minimum after t=0 is-
Solution
To solve this problem, we need to use the principles of physics, specifically those related to force, potential energy, and acceleration.
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First, we need to find the force acting on the block. The force is the negative derivative of the potential energy with respect to position (x). So, F = -dU/dx = 3x^2 - 12x.
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The acceleration of the block is the force divided by the mass. So, a = F/m = (3x^2 - 12x)/2.
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To find the minimum acceleration, we need to find the derivative of the acceleration with respect to x and set it equal to zero. So, da/dx = 3x - 6 = 0. Solving for x gives x = 2.
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The velocity of the block is the integral of the acceleration with respect to time. However, since we don't know the time when the acceleration is minimum, we can't directly integrate the acceleration to find the velocity.
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Instead, we can use the principle of conservation of energy. The total energy of the block (kinetic energy + potential energy) is conserved. So, 1/2mv^2 + U = constant.
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At t = 0, the block has velocity 45√ m/s and is at the origin, so its total energy is 1/22(45√)^2 + (-0^3 + 6*0^2 + 15) = 4050.
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When the acceleration is minimum (at x = 2), the total energy is 1/22v^2 + (-2^3 + 6*2^2 + 15) = 4050.
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Solving this equation for v gives v = sqrt((4050 - (-8 + 24 + 15))/2) = sqrt(4019) m/s.
So, the velocity of the block when its acceleration is minimum after t=0 is sqrt(4019) m/s.
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