Question

A block of $10\text{kg}$ is pulled in a vertical plane along a frictionless surface in the form of an arc of a circle of radius $10\text{m}$. A force of $200\text{N}$ is applied such that tension in the string always remains along the tangential direction as shown in figure. If the block started from rest at $A$, the velocity at $B$ would be:
Consider $\frac{\pi }{3}=1$ for calculation and take $g=10{\text{m/s}}^2$)

• A. $10\sqrt{3}\text{m/s}$
• B. $10\text{m/s}$
• C. $\frac{5}{\sqrt{3}}\text{m/s}$
• D. $3\sqrt{3}\text{m/s}$

Answer 1

The correct option is A $10\sqrt{3}\text{m/s}$

Tension will be equal to applied force $F=200\text{N}$ and its direction is always tangential to the circular arc. Angle between tension and displacement is $0^{\circ }$.
So, work done by tension
$W_T=\int Fdl\cos 0^{\circ }=\int Fdl$
$W_T=F\int dl=Fl...\left(i\right)$
where $l$ is the length of the arc.


$l=2\pi R\times \frac{\theta }{{360}^{\circ }}=2\pi \times 10\times \frac{{60}^{\circ }}{{360}^{\circ }}$
$\Rightarrow W_T=F\times \frac{10\pi }{3}=200\times \frac{10\pi }{3}$
Substituting given data $\frac{\pi }{3}=1$
$W_T=2000\text{J}$
Now, rise in height of the block is,
$h=R-R\cos {60}^{\circ }=10(1-\frac{1}{2})=5\text{m}$
$W_{mg}=-mgh=-10\times 10\times 5=-500\text{J}$
Work done by normal reaction, $W_N=0\text{since}N\perp dl$

Applying W.E.T on block from $A$ to $B$:
$W_{\text{ext}}=\Delta KE$
$W_T+W_{mg}+W_N=K{E}_f-K{E}_i$
$2000-500+0=\frac{1}{2}\times 10\times v^2-0$
$v^2=300$
$\therefore v=10\sqrt{3}\text{m/s}$