First slide

Work

Question

A bead of mass ½ kg starts form rest from A to move in a vertical plane along a smooth fixed quarter ring of radius 5 m, under the action of a constant horizontal force F= 5N as shown in the figure. The speed of bead as it reaches point B is

Moderate

A

$14.14 m s^{- 1}$
B

$7.07 m s^{- 1}$
C

$5 m^{- 1}$
D

$25 m^{- 1}$

Solution

Applying the work energy theorem we get
$\begin{array}{l} c h a n g e i n K E = W o r k d o n e b y h o r i z o n t a l f o r c e a n d g r a v i t a t i o n a l f o r c e \Rightarrow \frac{1}{2} \times m v^{2} - 0 = F \times R + m g \times R \\ s u b s t i t u t e t h e g i v e n v a l u e s \\ \frac{1}{2} \times \frac{1}{2} \times v^{2} = 5 \times 5 + \frac{1}{2} \times 10 \times 5 \Rightarrow v^{2} = 4 x 50 \\ v = \sqrt{200} = 14.14 m S^{- 1} \end{array}$

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