Question

Figure shows a ball of mass m connected with two ideal springs of force constant k, kept in equilibrium on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration of ball?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Well initially I can say that the ball was in equilibrium and hence the net force on the ball will be 0.

Okay

How many force you think are acting on the ball

(1)Mg vertically downward for sure as earth will attract it.

Since the ball is in contact with the incline and is pushing the incline down so the incline will push if up with a Normal and we know by Now that it will be perpendicular to the contact surface

Let's resolve mg along and perpendicular to incline.

So $\frac{mg}{\sqrt{2}}$ N in the assumed y - axis and there is a force $\left(\frac{mg}{\sqrt{2}}\right)$ in (-y axis) so this has to be balanced too.

Hmmm. Which force balances that

Oh !!!

There are 2 springs also

So there will be elongation in the spring. Lets say x so they both have kx force.

Okay, Lets see the top view

Since in equilibrium so you know

$\sum F_y=0$

$\Rightarrow \frac{2kX}{\sqrt{2}}-\frac{mg}{\sqrt{2}}=0$

Kx = $\frac{mg}{2}$ ----------(1)

Net force in y

$\sum F_y=\frac{kx}{\sqrt{2}}-\frac{mg}{\sqrt{2}}$

From eq. (i)

$\sum F_y=\frac{mg}{2\sqrt{2}}-\frac{mg}{\sqrt{2}}=-\frac{mg}{2\sqrt{2}}$

So $a_y=\frac{\sum F_y}{m}=\frac{-g}{2\sqrt{2}}$

Similarly

Net force is x direction

$\sum F_x=\frac{kx}{\sqrt{2}}$

From eq (1)

$\sum F_x=\left(\frac{-mg}{2\sqrt{2}}\right)$

$a_x=\frac{\sum F_x}{m}=\frac{-mg}{2\sqrt{2}}=\frac{-g}{2\sqrt{2}}$

$a_{net}=\sqrt{(a_x{)}^2+(a_y{)}^2}$

= $5m/s^2$