Question
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?
The answer is 172 cm/s; but I cannot figure out how to get this answer.
The answer is 172 cm/s; but I cannot figure out how to get this answer.
Answers
drwls
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.
(-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.
Annabelle
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?
What gravitational potential energy term are you referring to?