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Question Number 103792 by aurpeyz last updated on 17/Jul/20

	Answered by Dwaipayan Shikari last updated on 17/Jul/20

	$v_{f}^{2} = v_{0}^{2} + 2 a s$ $400 = - 2 a . 120$ $a = - \frac{400}{240} \frac{m}{s^{2}} = - \frac{5}{3} \frac{m}{s^{2}} (n e g a t i v e m e a n s d e a c c l e r a t i o n)$ $F o r c e o n p u c k = - m . \frac{5}{3} N$ $I s e q u a l t o f r i c t i o n a l f o r c e f = - μ m g (n e g a t i v e d e p e n d s o n s i g n c o n v e n t i o n$ $- m \frac{5}{3} = - μ m g$ $μ = \frac{5}{30} = \frac{1}{6} = 0.1666666$
	Answered by Dwaipayan Shikari last updated on 17/Jul/20
	$(1/2)mv^2 =f.d {Change in kinetic energy=(1/2)m(v^2 −0)=(1/2)mv^2 (1/2)mv^2 =μmgd {Work done by the frictional force=f.d μ=(v^2 /(2gd))=((400)/(2400))=(1/6)=0.16666666..$
	$\frac{1}{2} {m v}^{2} = f . d {C h a n g e i n k i n e t i c e n e r g y = \frac{1}{2} m (v^{2} - 0) = \frac{1}{2} {m v}^{2}$ $\frac{1}{2} {m v}^{2} = μ m g d {W o r k d o n e b y t h e f r i c t i o n a l f o r c e = f . d$ $μ = \frac{v^{2}}{2 g d} = \frac{400}{2400} = \frac{1}{6} = 0.16666666 . .$
	Commented by aurpeyz last updated on 17/Jul/20

	$thank you Sir . is there any method apart from conservstion of energy that can be used ?$
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