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Potential Energy

In a soring g...

Question

In a soring gun having a spring constant 100N/m, a ball of mass 0.1 kg is put in its barrel by compressing the spring through a distance of 0.05m.
If the ball leaves the gun horizontally at a height of 2m above the ground.
1)Find the velocity of the bulb when the spring is released.
2)Where should be a box be placed on the ground so that the ball falls in it?

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Solution

$P o t e n t i a l e n e r g y s t o r e d i n t h e s p r i n g = \frac{1}{2} k x^{2} P E = \frac{1}{2} \times 100 \times 0 . 05^{2} P E = 0.125 J t h i s i s t h e a m o u n t o f e n e r g y t h a t w i l l b e p r o v i d e d t o t h e b a l l i n t h e f o r m o f k i n e t i c e n e r g y . K E o f b a l l = 0.125 J \frac{1}{2} m v^{2} = 0.125 v^{2} = \frac{0.125 \times 2}{0.1} v = 1.58 m s^{- 1} (A N S W E R I) f o r c e e x e r t e d o n t h e b a l l = f o r c e e x e r t e d o n t h e s p r i n g = k x F_{b a l l} = 100 \times 0.05 F_{b a l l} = 5 N a_{b a l l} = \frac{F_{b a l l}}{m_{b a l l}} = \frac{5}{0.1} = 50 m s^{- 2} n o w, T i m e t a k e n b y t h e b a l l t o f a l l 2 m l e t i t b e' t' t h e n, 2 = u t + \frac{1}{2} g t^{2} u = 0 \sin c e n o i n i t i a l v e r t i c a l v e l o c i t y 2 = \frac{10}{2} t^{2} 0.4 = t^{2} t = 0.63 s n o w, f o r h o r i z o n t a l m o t i o n u = 1.58 m s^{- 1} t = 0.63 s a n d, a = 50 m s^{- 2} s = u t + \frac{1}{2} a t^{2} s = 1.58 \times 0.63 + \frac{50 \times 0 . 63^{2}}{2} s = 10.92 m (A N S W E R I I)$

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