1. Tardigrade
  2. Question
  3. Physics
  4. 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 vecF=(4-x2) veci N, where x is in meter and the initial position of the block is x=0. The maximum positive displacement x is

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Q. 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 $\vec{F}=\left(4-x^{2}\right) \vec{i} N$, where $x$ is in meter and the initial position of the block is $x=0$.
The maximum positive displacement $x$ is

Work, Energy and Power

Solution:

KE of the particle at $x=x$ is $K=4 x-\frac{x^{3}}{3}$
At maximum displacement, velocity will be zero or
$K=0$,i.e., $x=2 \sqrt{3} m$