Question

PART 1/4
A massless spring with force constant
502 N/m is fastened at its left end to a vertical
wall, as shown below.
(PICTURE 1)
The acceleration of gravity is 9.8 m/s^2
Initially, the 8 kg block and 4 kg block rest
on a horizontal surface with the 8 kg block in
contact with the spring (but not compressing it) and with the 4 kg block in contact with the
8 kg block. The 8 kg block is then moved to
the left, compressing the spring a distance of
0.2 m, and held in place while the 4 kg block
remains at rest as shown below.
(PICTURE TWO)
Determine the elastic energy U stored in
the compressed spring.
Answer in units of J.

PART 2/4
The 8 kg block is then released and accelerates
to the right, toward the 4 kg block. The
surface is rough and the coefficient of friction
between each block and the surface is 0.5 . The
two blocks collide, stick together, and move
to the right. Remember that the spring is not
attached to the 8 kg block.
Find the speed of the 8 kg block just before
it collides with the 4 kg block.
Answer in units of m/s.

PART 3/4
Find the final speed of both blocks (stuck
together) just after they collide.
Answer in units of m/s.

PART 4/4
Find the horizontal distance the blocks move
before coming to rest.
Answer in units of m.

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