Question

A 77.3 g mass is attached to a horizontal spring with a spring constant of 12.5 N/m and released from rest with an amplitude of 38.9 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless? Answer in units of m/s.

Answer 1

Answer:

speed of the mass is 3.546106 m / s

Explanation:

given data

mass = 77.3 g = 77.3 × $10^{-3}$  kg

spring constant k = 12.5 N/m

amplitude A = 38.9 cm = 38.9 ×$10^{-2}$ m

to find out

the speed of the mass

solution

we will apply here conservation energy  that is

K.E + P.E = Total energy  ..................1

so that Total energy = K.E max = P.E max

we know  amplitude so we find out first P.E max that is  

PE max = K.E + P.E  

(1/2)kA² = (1/2)mv² + (1/2)kx²  

kA^² =  mv²+ kx²

so here v²  will be

v²  = k(A² - x²) / m  

v = √[(k/m)×(A² - x²)]  ............2

here x = (1/2)A   so from from 2 equation

v = √[(k/m)×(A² - (A/2)²)]

v = √[(k/m)×(3/4×A²)]

now put all value

v = √[(12.5/ 77.3 × $10^{-3}$ )×(3/4×(38.9 ×$10^{-2}$)²)]

v = 3.546106 m / s

speed of the mass is 3.546106 m / s