A 100-g block hangs from a spring with k = 5.2 N/m. At t = 0 s, the block is 20.0 cm below the equilibrium position and moving upward with a speed of 216 cm/s. What is the block's speed when the displacement from equilibrium is 27.0 cm?

Question

The answer is 172 cm/s; but I cannot figure out how to get this answer.

Answer 1

You need to consider gravitational potential energy, spring potential energy, and block kinetic energy. The sum of the three is constant. I will refer gravitational P.E. to the equilibrium position

(-0.2 m) M*g + (1/2)k(0.2)^2 + (M/2)(2.16 m/s)^2
= (-0.27 m)M*g + (1/2)k(0.27)^2 + (M/2)V^2

(M/2)V^2 = 0.0686 J + 0.2333 J - 0.0855 J = 0.2164 J

V^2 = 4.328 m^2/s^2
V = 2.08 m/s

In order to get the 1.72 m/s answer, you would have to negelect the gravitational potential energy term. I don't agree with doing that.

Answer 2

When I follow your steps I get 1284 cm/s, so I'm not sure how you're getting 2.08 m/s.

What gravitational potential energy term are you referring to?