First slide

Conservation of mechanical energy

Question

An ideal massless spring S can be compressed 1 m by a force of 100N in equilibrium .The same spring is placed at the bottom of a frictionless inclined plane at $30^{o}$ to the horizontal. A 10kg block M is releases from rest at the top of the incline and is brought to rest momentarily after compressing the spring by 2m. If g = 10 $m s^{- 2}$ , what is the speed of mass just before it touches the spring?

Moderate

A

$\sqrt{20} m s^{- 1}$
B

$\sqrt{30} m s^{- 1}$
C

$\sqrt{10} m s^{- 1}$
D

$\sqrt{40} m s^{- 1}$

Solution

$\begin{array}{l} r e s t o r i n g f o r c e o f a s p r i n g i s F = - k x \\ ∴ s p r i n g c o n s \tan t i s k = \frac{F}{x} = \frac{100}{1} = 100 N / m \\ a c c o r d i n g t o p r i n c i p l e o f c o n s e r v a t i o n o f e n e r g y, w h e n m a s s = 10 k g s l i d e d o w n i n c l i n e \\ E_{i n i t i a l} = E_{f i n a l} \\ K E_{M a s s} + P E_{M a s s} = e n e r g y s t o r e d i n a s p r i n g \\ ∴ \frac{1}{2} \times 10 \times v^{2} + (10) (10) (2 \sin 30^{0}) = \frac{1}{2} \times 100 \times {(2)}^{2} \\ s o l v i n g w e g e t \\ v e l o c i t y o f m a s s i s V = \sqrt{20} m / s \end{array}$

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