Respuesta :
Answer:
amplitude is 8.92 cm
speed of block is 44.11 cm/s
Explanation:
given data
mass = 0.650 kg
spring constant = 18 N/m
speed = 47 cm/s = 0.44 m/s
speed at x point = 0.350 A
to find out
amplitude of subsequent oscillation
solution
we know here conservation of energy
maximum kinetic energy = maximum potential energy
[tex]\frac{1}{2} m v^{2} = \frac{1}{2} k A^{2}[/tex]
here we know k is spring constant and m is mass and A is amplitude and v is velocity
so solve it we get
A = [tex]\sqrt{\frac{mv^{2} }{k} }[/tex]
put here all these value
A = [tex]\sqrt{\frac{0.650(0.47)^{2} }{18} }[/tex]
A = 0.08931 m
so amplitude is 8.92 cm
and
by conservation of energy
initial energy = final energy
[tex]\frac{1}{2} m V^{2} + \frac{1}{2} k x^{2}[/tex] = [tex]\frac{1}{2} m Vm^{2} + 0 [/tex]
solve we gey V
V = [tex]\sqrt{Vm^{2} - \frac{k }{m} x^{2} }[/tex]
put here value Vm = 0.47 , and x = 0.350
V = [tex]\sqrt{0.47^{2} - \frac{18 }{0.650} (0.350*0.088 m)^{2} }[/tex]
V = 0.4411 m/s
so speed of block is 44.11 cm/s
