Answer:
1. 10.0 J
2. 0.742 m/s
3. 0.494 m/s
4. 0.0249 m
Explanation:
(1/4) The elastic energy in a spring is:
EE = ½ k x²
EE = ½ (502 N/m) (0.2 m)²
EE ≈ 10.0 J
(2/4) The energy in the spring is converted to kinetic energy in the block and work by friction.
EE = KE + W
EE = ½ m v² + Fd
10.0 J = ½ (8 kg) v² + (8 kg × 9.8 m/s² × 0.5) (0.2 m)
v ≈ 0.742 m/s
(3/4) Momentum is conserved.
Momentum before = momentum after
(8 kg) (0.742 m/s) = (8 kg + 4 kg) v
v ≈ 0.494 m/s
(4/4) The kinetic energy of the blocks is converted to work by friction.
KE = W
½ m v² = Fd
½ (8 kg + 4 kg) (0.494 m/s)² = ((8 kg + 4 kg) × 9.8 m/s² × 0.5) d
d ≈ 0.0249 m
