A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of x-axis is applied to the block. The force is given by $\to F = (4 - x^{2})^i$ N, where x is in meter and the initial position of the block is x=0.The maximum positive displacement x is

2 \sqrt{3}

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2 m

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4 m

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\sqrt{4}

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Solution

The correct option is A $2 \sqrt{3}$ m
Maximum booger $= [9 x - \frac{x^{3}}{3}]$
$\Rightarrow 4 x - \frac{x^{3}}{3} = 0$
$\Rightarrow x^{3} = 4 x + 3$
$\Rightarrow x^{2} = 12$
$\Rightarrow x = 2 \sqrt{3} .$
Hence, the answer is $2 \sqrt{3} m .$

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Similar questions

Q. A

1.5 k g

block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of the x-axis is applied to the block. The force is given by

\to F = (4 - x^{2})^i

N, where

x

is in meter and the initial position of the block is

x = 0

.The maximum kinetic energy of the block between

x = 0

and

x = 2.0

m is

Q. A block of mass 2 kg is initially at rest on a horizontal frictionless surface. A horizontal force

\to F = (9 - x^{2})^i

N acts on it, when the block is at

x = 0

. The maximum kinetic energy of the block between

x = 0

and

x = 3

m in Joule is:

A 100 gm block is connected to a horizontal massless spring of force constant 25.6 N/m and is free to oscillate on a horizontal frictionless surface. The block is displaced on a horizontal frictionless surface. The block is displaced by 3 cm along positive x - direction and released with an initial velocity of $v_{i} = - 16 \sqrt{3}$ cm.

Express position of block as function of time.

A block of mass 3 kg in contact with a second block of mass 2 kg rests on a horizontal frictionless surface . A horizontal force of 10 N is applied to push the first block. The force with which the first block pushes the second block is:

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The correct option is A 2√3 mMaximum booger =[9x−x33]⇒4x−x33=0⇒x3=4x+3⇒x2=12⇒x=2√3.Hence, the answer is 2√3m.

The correct option is A $2 \sqrt{3}$ m
Maximum booger $= [9 x - \frac{x^{3}}{3}]$
$\Rightarrow 4 x - \frac{x^{3}}{3} = 0$
$\Rightarrow x^{3} = 4 x + 3$
$\Rightarrow x^{2} = 12$
$\Rightarrow x = 2 \sqrt{3} .$
Hence, the answer is $2 \sqrt{3} m .$