Question

Block A of 15kg of the figure below slides on a surface where the coefficient of kinetic friction is uk=0.3. The block moves at 10m/s when it is s=4m from a second 10kg B block, initially at rest. Considering this situation, determine what will be the maximum compression of the spring, initially decompressed, after the collision if the coefficient of return between the blocks is e=0.6. Tip here, as the blocks are subject to friction forces, use the equation that relates mechanical energy and the work done by non-conservative forces.

Answer 1

0.287 m

Step-by-step explanation:

The velocity of block A when it reaches block B is:

KE₀ = KE + W

½ mv₀² = ½ mv² + Fd

½ mv₀² = ½ mv² + mg μ d

v₀² = v² + 2g μ d

v² = v₀² − 2g μ d

v² = (10 m/s)² − 2 (9.8 m/s²) (0.3) (4 m)

v = 8.75 m/s

The coefficient of restitution is 0.6, so the relative velocity between A and B after the collision is:

e = |Δv after| / |Δv before|

0.6 = Δv / (8.75 m/s)

Δv = 5.25 m/s

Momentum is conserved, so the speed of block B after the collision is:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(15 kg) (8.75 m/s) + (10 kg) (0 m/s) = (15 kg) (v) + (10 kg) (v + 5.25 m/s)

131.2 m/s = 25v + 52.5 m/s

25v = 78.7 m/s

v = 3.15 m/s

Energy is conserved, so the compression of the spring is:

KE = EE + W

½ mv² = ½ kd² + Fd

½ mv² = ½ kd² + mg μ d

½ (10 kg) (3.15 m/s)² = ½ (1000 N/m) d² + (10 kg) (9.8 m/s²) (0.3) d

49.6 = 500 d² + 29.4 d

0 = 500 d² + 29.4 d − 49.6

d = [ -29.4 ± √(29.4² − 4(500)(-49.6)) ] / 1000

d = (-29.4 ± 316.2) / 1000

d = 0.287 m