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Laws of Motion — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion1 Mark
MCQ
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 $\overrightarrow{ F }=\left(4- x ^2\right) \hat{ i } N$,where x is in metre 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
  • A
    2.33 J
  • B
    3.33 J
  • ✓
    5.33 J
  • D
    6.67 J
✓

Answer

Correct option: C.
5.33 J
(C)
From work-energy theorem, kinetic energy of block at $x =0$ to x is;
$K=\int_0^x\left(4-x^2\right) \cdot d x$
$\therefore \quad K =4 x -\frac{ x ^3}{3}$
For K to be maximum, $\frac{ dK }{ dx }=0$
or $4-x^2=0$
or $x= \pm 2 m$
At $x=+2 m, \frac{d^2 K}{d x^2}$ is negative.
i.e., kinetic energy ( K ) is maximum.
$\therefore \quad K _{\max }=(4)(2)-\frac{(2)^3}{3}=\frac{16}{3} J=5.33 J$

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