Question

$\huge{\underline{\tt{Ello!!}}}$
❄️ A block whose nass is 2 kg is fastened to a spring. The spring has a constant of 50 N/m. The block is pulled to a distance x= 10 cm from its equilibrium postion at x=0 on a frictionless surface from rest rest at t= 0. Calculate the. kinetic, potential and total energies of the block when it is 5 cm away from the mean postion.
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Answer 1

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Mass= 2kg

Spring constant = 50N/m

Distance x=10cm

X=0

t=0

Kinetic and potential energy of the block = 5 cm away from the mean position

$angular \: frequency \: (w) = \sqrt{k \div m} \\ w = \sqrt{100 \div 2} \\ w = \sqrt{50} \\ w = 7.071rads {}^{-}$

$Its \: displacement \: at \: any \: time(t) \: is \: given \: by \: x(t) = (0.1)cos(7.07t) \\ pe = 1 \div 2kx {}^{2} \\ = 1 \div 2(50)(0.1) {}^{2} \\ = 25 \times 0.01 \\ = 0.25j$

Answer 2

Answer:

donno....sorry

Explanation:

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