Lease answer the question and show all work clearly and easy to read. So that I

Question

Lease answer the question and show all work clearly and easy to read. So that I can understand how to do the problem. Thank you. 2. A mass m = 2 kg hangs from a light vertical spring, of spring constant k-80 N/m. attached to the ceiling. Without the mass, the bottom of spring hangs at a height y-0 m. For this problem you should use g 10 s for easier calculation (a) Find y the height that the mass hangs at on the spring when at equilibrium 3 pts (b) Now the mass is pulled down further to a lower point y-0.6 m and released. Find ys, the maximum height the mass reaches. (Ignore any effects of friction or air resistance and assume the mass can only move along a line in the vertical direction.) [5 pts

Explanation / Answer

a) At equillibrium the net force will be zero.

There are two forces working on the block, force due to gravity and force due to spring.

Hence the spring force will be equal to the force due to gravity.

Fg = mg = ky_1

hence y_1 = mg/k = 2*9.81/80 = 24.525 cm

b)

y2 = -60 cm

The amplitude of SHM will be A = |y2-y1| = 35.475 cm

hence the maximum height will be y3 = |A-y1| = 35.475-24.525 = 10.95 cm.

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