A block with mass m = 4.4 kg is attached to two springs with spring constants kl

Question

A block with mass m = 4.4 kg is attached to two springs with spring constants kleft = 35.0 N/m and kright = 49.0 N/m. The block is pulled a distance x = 0.24 m to the left of its equilibrium position and released from rest. -Where is the block located, relative to equilibrium, at a time 0.88 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value) -What is the net force on the block at this time 0.88 s? (a negative force is to the left; a positive force is to the right)

Explanation / Answer

In fact the springs are in parallel, though seems to be in series. The extreme ends are fixed and the block is connected to the other ends of each spring ---------------------------------------… v² = (k1+ k2)*x²/m v² = (33+ 56)*0.2²/5.5 v = 0.81m/s to ward right. -------------------------------- ? = v (k1+ K2)/m =v [(33+ 56)*/5.5] = 4.02rad/s x = A sin (?t –p/2) = 0.2 sin (4.02t - p/2) (angles are in radians) At time t = 0.94s x = 0.161m to the right of the equilibrium position, ------------------------------- F = -kx = - (33+ 56)*(0.161) = - 14.329toward left ---------------------------------- Total energy =0.5*(33+ 56)*0.2² = 1.78 J --------------------------------------- Edited since there was a mistake in calculating ?

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