Question

A 1.5 kg of block is initially at rest on a
horizontal frictionless surface when a
horizontal force in the positive direction of
X-axis is applied on the block. The force is
F=(4-x?)in, where x is in metre and
initially the block is at x = 0. The maximum
KE of the block is
(1) 2.33 J
(2) 8.67 J
(3) 5.33 J
(4) 6.67 J​

Answer 1

Answer:

From work energy theorem

Kinetic energy of block

$k=\int \limits_0^x (4-x^2)dx$

$k=4x-\large\frac{x^3}{3}$

For K to be maximum $\large\frac{dh}{dx}$$=0$

$\large\frac{dk}{dx}$$=4-x^2$

$4=x^2$

$x=\pm 2$

At $x=+2 \quad \large\frac{d^2k}{dx^2}$ is -ve . So k is maximum

$k_{max}=4(2)-\large\frac{(2)^3}{3}=\large\frac{16}{3}$$J=5.33\;J$

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