Answer:
(a) The total energy of the spring system is 0.032 J
(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s
(c) The kinetic energy when its position is 2.50 cm is 0.0195 J
Explanation:
The given parameters are;
The mass of the object connected to the spring, m = 35.0 g = 0.00
The force constant, k = 40.0 N/m
The amplitude of the oscillation, a = 4.00 cm = 0.04 m
Therefore, we have
(a) The total energy of the spring system, E given as follows;
E = PE + KE = 1/2·m·v² + 1/2·k·x²
Where;
v = The velocity of the spring
x = The extension of the spring
When the spring is completely extended, x = a, and v = 0, therefore;
The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J
(b) At x = 1.20 cm = 0.012 m, we have;
E = 1/2·m·v² + 1/2·k·x²
0.032 = 1/2 × 0.035 × v² + 1/2 × 40 × 0.012²
0.032 - 1/2 × 40 × 0.012² = 1/2 × 0.035 × v²
0.02912 = 1/2 × 0.035 × v²
1/2 × 0.035 × v² = 0.02912
v² = 0.02912/(1/2 × 0.035) = 1.664
v = √1.664 ≈ 1.28996
The speed of the object when its position is 1.20 cm, v ≈ 1.28996 m/s
(c) When its position is 2.50 cm = 0.025 m, we have;
E = PE + KE
0.032 = 1/2 × 40 × 0.025² + KE
KE = 0.032 - 1/2 × 40 × 0.025² = 0.0195
The kinetic energy when its position is 2.50 cm = 0.0195 J.
