Question

↵An object starts from rest and is acted upon by a variable force F as shown in figure. If F0 is the initial value of the force, than the position of the object, where it again comes to rest will be


2F0​ / tanα

F0​ / sinα

2F0​ / cotα

F0​ / 2 cosα

Answer 1

here in figure, we can see that force is function of x , graph between force and displacement is straight line with intersects at -Fo on y-axis.
so, $F_{net}=F(x)$

we know from Newton's law of motion,
Fnet = ma = $mv\frac{dv}{dx}$
so, $mv\frac{dv}{dx}=F(x)$
but $F(x)=(tan\alpha)x-F_0$
$\implies m\int\limits^v_0{v}\,dv=\int\limits^x_0\{(tan\alpha)x-F_0\}\,dx$
$\implies m\frac{v^2}{2}=(tan\alpha)\frac{x^2}{2}-F_0x$

for again come to rest , v = 0
then, $tan\alpha\frac{x^2}{2}-F_0x=0$
$x=\frac{2F_0}{tan\alpha}$

hence, option (A) is correct.

Answer 2

Ans

Explanation:tanx=f°/X

X=f°/tanx

Final &initial K.E =0

Work=O

Area should be 0

Therefore the X should be equal in both negative and positive work done graph...x°(total)= 2f/tanx