Question

I didn't understand the previous person's hand writing, can someone else solve it please, thank you.

A spring stress 0.150 m when a 0.300 kg mass is gently suspended from it as fig below. The spring is then set up horizontally with 0.300 kg mass resting on a frictionless table. The mass is pulled so that the spring is stretched 0.100 m from the equilibrium point, and released from rest. Determine: a) the spring stiffness constant K; b) the amplitude of the horizontal oscillation A; c) the magnitude of the maximum velocity Vmax? d) the magnitude of the velocity V when the mass is 0.075 m from the equilibrium; e) the total energy;

Answer 1

a)spring stretched by x=0.15 m

suspended mass = 0.3 kg

F = kx

force F = 0.3g = 2.94 N

k = F/x = 2.94/0.15 = 19.6 N/m

b) horizontal motion.

spring pulled by 0.1 m

amplitude of the oscillations A = 0.1 m

c) max. vel:

The velocity is maximum when the spring is at its normal length

The spring energy is max. when it is fully stretched x=0.1 m

U = 1/2*kx² = k/2*(0.1)²

Total potential energy of the spring kx²/2 is equal to the KE of the mass 1/2 mv²  

k(0.1)² = mv²

v_max = 0.1(19.6/0.3)^1/2 = 0.81 m/s

d)

mass is at x= 0.75 m from the equilibrium

spring energy = 1/2 k(0.75)²  

KE of the mass = 1/2 mv² = 1/2 k(0.1)² - 1/2 k(0.075)²  

v(x=0.075)= 0.12 m/s

e)Total energy = 1/2 mv_max² = 1/2 kx_max² = 1/2*19.6*0.1² = 0.098 J