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Question Number 121434 by mathdave last updated on 08/Nov/20

Commented by mathdave last updated on 08/Nov/20

$a h e l p f o m a n y b o d y p l s$
Commented by Tawa11 last updated on 06/Sep/21

$great$
	Answered by mr W last updated on 08/Nov/20

	$s p e e d o f b l o c k b e f o r e r e a c h i n g r o u g h$ $s u r f a c e = V_{1}$ $\frac{1}{2} {m V}_{1}^{2} + m g h = \frac{1}{2} {m V}_{0}^{2}$ $w i t h h = h e i g h t o f r a m p = l_{r a m p} \sin θ$ $V_{1}^{2} = V_{0}^{2} - 2 {g l}_{r a m p} \sin θ$ $\Rightarrow V_{1} = \sqrt{V_{0}^{2} - 2 {g l}_{r a m p} \sin θ}$ $= \sqrt{20^{2} - 2 \times 10 \times 2 \times \sin 30 °} = 2 \sqrt{95} m / s$ $= 19.45 m / s$ $s p e e d o f b l o c k a f t e r l e a v i n g r o u g h$ $s u r f a c e = V_{2} = b e f o r e h i t t i n g s p r i n g$ $k i n e t i c e n e r g y :$ $E_{k} = \frac{1}{2} {m V}_{2}^{2} = \frac{1}{2} {m V}_{1}^{2} - μ_{k} {m g l}_{r o u g h}$ $= \frac{1}{2} {m V}_{0}^{2} - {m g l}_{r a m p} \sin θ - μ_{k} {m g l}_{r o u g h}$ $= m [\frac{1}{2} V_{0}^{2} - g (l_{r a m p} \sin θ + μ_{k} l_{r o u g h})]$ $= 5 [\frac{1}{2} \times 20^{2} - 10 (2 \times \sin 30 ° + 0.4 \times 15)]$ $= 650 J$ $\frac{1}{2} {K d}^{2} = E_{k}$ $\Rightarrow K = \frac{2 \times 650}{2^{2}} = 325 N / m$
	Commented by Lordose last updated on 08/Nov/20

	$Nice sir ▴$
	Commented by peter frank last updated on 08/Nov/20

	$thank you$
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