Respuesta :
Answer:
a) 2816.04 N/m b) 55.2 J c) 9.3 kg d) 2.03 m/s e) 19.15 J f) 36.05 J g) 0.166 m
Explanation:
Total mechanical energy = 55.5 J, maximum displacement of the block (Xo) = 0.198 m
a) total energy = kinetic energy + potential energy where at maximum displacement the kinetic energy =0 since the body temporarily stop
55.2 = 1/2 K (Xo)² where Xo = 0.198 m
55.2 × 2 = K (0.198²)
K = (55.2 × 2) / (0.198²) = 2816.04 N/m
b) at the equilibrium position since the surface is frictionless
Kinetic energy = total mechanical energy = 55.2 J
c) Kinetic energy = 1/2mVmax²
55.2 = 1/2m Vmax² where Vmax = 3.45 m/s
55.2 × 2 = m × (3.45²)
m = (55.2 × 2) / (3.45²) = 9.3 kg
d) total energy = kinetic energy + potential energy
total energy = 1/2mv² + 1/2K x² where v is the speed at 0.160m and x is the displacement from equilibrium point which = 0.16m
55.2 = 1/2×9.3 × v² + 1/2 × 2816.04 × (0.16²)
55.2 = 1/2×9.3v² + 36.05
55.2 - 36.05 = 1/2 × 9.3v²
19.15 × 2 / 9.3 = v²
v² = 4.12
v = √4.12 = 2.03 m/s
e) kinetic energy at 0.016m = 1/2 mv² = 1/2 ×9.3 × (2.03²) = 19.15 J
f) potential energy = 1/2 K x² = 1/2 × 2816.04 × (0.16²) = 36.05 J
g) total energy = 55.2 J at 0.198 m where kinetic energy at this point = 0 since the body was at maximum displacement from it equilibrium point
when the body was released its loss 16.6 to the first turning point after passing equilibrium at x = 0
at the turning point energy is conversed, kinetic energy at the point = 0
total energy = 55.2 - 16.6 energy loss to friction
total energy left = 38.6 = 1/2 × K × x1² ( unkown new maximum displacement)
38.6 × 2/ 2816.04 = x1²
x1² = 0.0274
x1 = √0.0274 = 0.166 m
