Question

PART 1/4

Answer 1

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