A horizontal spring attached to a wall has a force constant of 820 N/m. A block

Question

A horizontal spring attached to a wall has a force constant of 820 N/m. A block of mass 1.20 kg is attached to the spring and oscillates freely on a horizontal, frictionless surface as in the figure below. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? This answer has not been graded yet. (b) What are the two points of interest? This answer has not been graded yet. (c) Find the energy stored in the spring when the mass is stretched 5.60 cm from equilibrium and again when the mass passes through equilibrium after being released from rest. x = 5.60 cm J x = 0 cm Correct: Your answer is correct. J (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. m/s (e) What is the speed at the halfway point? m/s Why isn't it half the speed at equilibrium?

Explanation / Answer

(a) Energy = (1/2)KX2 + (1/2)mV2
dE/dt = 0
KXV + (mVa) = 0
a + K/m X = 0
therefore natural frequency is
w = (K/m)1/2
w = (820/1.2)1/2 = 26.14 rad/s
(a) Spring and mass constitute the system, spring force that is Kx this system is interacting
(b)Extreme points are the point of interest
(c)Energy stored in the spring = (1/2)KX2 = (1/2)(820)*(0.056)2 = 1.28576 J
(d) Eenrgy = (1/2)KX2 + (1/2)Mv2
where M is mass ,V is velocity , K is spring constant and X is the distance by which spring is stretched.

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