When a 13.8-kg block is attached to the free end of a vertically hanging spring, he spring stretches by 21 cm. Wjat would the mass of the block have to be in order for the spring to stretch by 31.0 cm?
Given
mass, m= 13.8 kg
x = 21 cm = 0.21 m
a) since F = kx
mg = kx
13.8 kg * 9.8 m/s^2 = k * 0.21 m
k = 644 N/m
b) period , T = 2π√m/k
T = 2 * 3.14 *√13.8 kg/644 N/m
T = 0.91 s
c)Frequency , f = 1/T = 1/0.91s
f = 1.09 Hz
d)The amplitude is the maximum displacement from the equilibrium
position . so A = 31cm
e)Maximum speed , v = Aω = A(2π/T)
= (0.31 m) (2 *3.14)/0.91s
v = 2.13 m/s
Get Answers For Free
Most questions answered within 1 hours.