The potential energy stored in the wound up spring of a toy motor car is 18J. Ignoring any frictional resisitance , find the distance this car(mass 200g) will travel up a slope of 1 in 5

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Solution

$S p r i n g P o t e n t i a l e n e g y = 18 J S O i t c a n g o t i l l a h e i g h t o f = m g h = 18 s o h = 18 / m g = 18 / 0.2 (9.8) = 9.18 m S l o p e = T a n θ = 1 / 5 = 0.2 T a n θ = 0.2 = 9.18 m / B a s e B a s e = 45.9 m S o t h e d i s \tan c e t r a v e l l e d = \sqrt{(9.18)^{2} + (45.9)^{2}} = \sqrt{(9.18)^{2} + (45.9)^{2}} = \sqrt{2191.0824} = 46.8 m$

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The potential energy stored in the wound up spring of a toy motor car is 18J. Ignoring any frictional resisitance , find the distance this car(mass 200g) will travel up a slope of 1 in 5

Spring Potential enegy = 18 J SO it can go till aheight of = mgh = 18 so h = 18 /mg = 18 /0.2(9.8)= 9.18 mSlope= Tan θ= 1/5 = 0.2Tan θ = 0.2 = 9.18 m/Base Base = 45.9 m So the distance travelled = (9.18)2+(45.9 )2= (9.18)2+(45.9 )2= 2191.0824= 46.8m