Question

Q. An object of mass 500g, initially at rest acted upon by a variable force where x component varies with x in the manner shown. The velocities of the object at point x = 8 m and x =12 m, would be the respective values of (nearly)

Answer 1

The velocity at point x = 8 is 23 m/s

The velocity at point x = 12 is 21 m/s

Explanation:

The figure is attached.

The object is initially at rest.

The graph under the force-displacement curve = Work done

work done from x=0 to x=8 = Change in kinetic energy from x = 0 to x = 8

If velocity at x = 8 be $v_1$

Then,

Area under the curve = $\frac{1}{2}mv_1^2-0$

$\implies \frac{1}{2}mv_1^2 = 20\times 5+10\times 3$

$\implies \frac{1}{2}\times 0.5\times v_1^2=100+30$

$\implies \frac{v_1^2}{4}=130$

$\implies v_1=2\times\sqrt{130}$

$\implies v_1=2\times 11.2$ m/s

$\implies v_1=22.8$ m/s

or, $v_1= 23$ m/s (approximate)

Similarly work done from x = 8 to x = 12 m = Change in kinetic energy

$\implies \frac{1}{2}m(v_2^2-v_1^2)=-20.5\times 2-\frac{1}{2}\times 0.5\times 2+10\times 2$

$\implies \frac{0.5}{2}(v_2^2-23^2)=-41-0.5+20$

$\implies \frac{v_2^2-22.8^2}{4}=-21.5$

$\implies v_2^2-22.8^2=-86$

$\implies v_2^2=-86+519.84$

$\implies v_2^2=433.84$

$\implies v_2=\sqrt{433.84}$

$\implies v_2=20.82$ m/s

Hope this answer is helpful.

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Answer 2

Explanation:

answer is option -d 23m/s and 20.6m/s