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Work

Question

A Block of mass m = 0.1kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching the distance $\frac{x}{2}$ from equilibrium position , it hits another block and comes to rest momentarily, while the other block moves with a velocity 3m/s. The total initial Energy of the spring is,

Moderate

A

1.5 J
B

0.6 J
C

0.3 J
D

0.8 J

Solution

$T h e b l o c k c o m e s t o r e s t a f t e r c o l l i s i o n \Rightarrow b l o c k v e l o c i t y b e f o r e c o l l i s i o n w a s 3 m / s, b e f o r e c o l l i s i o n, K E o f t h e s p r i n g b l o c k s y s t e m i s, K . E = \frac{1}{2} m v^{2} s u b s t i t u t e g i v e n v a l u e s \Rightarrow K E = \frac{1}{2} 0.1 \times 3^{2} \Rightarrow K E = 0.45 J o u l e - - - (1)$
$P . E . = \frac{1}{2} {kx}^{2} a t d i s p l a c e m e n t = \frac{x}{2}, P E = \frac{1}{2} k(\frac{x}{2})^{2} ---(2) T o t a l e n e r g y = \frac{1}{2} {kA}^{2} a t d i s p l a c e m e n t x, T E = \frac{1}{2} {kx}^{2} ---(3) s u b s t i t u t e e q u a t i o n (3) i n e q u a t i o n (2) P E = \frac{1}{4} T E t h e n r e m a i n i n g e n e r g y w i l l b e K E K E = \frac{3}{4} T E s u b s t i t u t e e q u a t i o n (1) 0.45 = \frac{3}{4} T E T E = 0.45 X \frac{4}{3} T o t a l e n e r g y = 0.6 J o u l e$
$∴ T . E = \frac{4}{3} \times 0.45 = 0.6 J$

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