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8-kg lead brick falls from a height of 1.6 m.
(a) Find its momentum as it reaches the ground.
Ikg-m/s
(b) What Impulse is needed to bring the brick to rest?
N-s
(c) The brick falls onto a carpet, 2.0 cm thick. Assuming the force stopping it is constant, find the average force the carpet exerts on the brick
N
(d) If the brick falls onto a 6,0 cm foam rubber pad, what constant force is needed to bring it to rest?

Respuesta :

The magnitude of the impulse needed to bring the brick to rest is given by the initial momentum of the brick.

The correct responses are;

  • (a) 44.8 kg·m/s
  • (b) 44.8 N·s
  • (c) 6,278.4 N
  • (d) 2,092.8 N

Reasons:

The velocity of the brick as it reaches the ground is given by the formula;

v² = 2·g·h

v = √(2·g·h)

Where;

h = The height of the brick, h = 1.6 m

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

v ≈ √(2 × 9.81 m/s² × 1.6 m) ≈ 5.6 m/s

(a) Momentum = Mass × Velocity

The momentum of the brick = 5.6 m/s × 8 kg = 44.8 kg·m/s

(b) Impulse = Force × Δt = Mass × Δv

Mass × Δv = 8 kg × (5.6 m/s - 0) = 44.8 kg·m/s = 44.8 N·s

Therefore, the impulse = 44.8 N·s

(c) The distance over which the carpet stops the brick = 2.0 cm = 0.02 m

Initial energy in the brick, E = m·g·h

Which gives;

E = 8 kg × 9.81 m/s² × 1.6 m = 125.568 J

Energy = Force × Distance

[tex]\displaystyle Force \ carpet \ exerts \ on \ brick, \ F = \frac{125.568 \, J}{0.02 \, m} = \mathbf{ 6,278.4 \, N}[/tex]

The average force the carpet exerts, F = 6,278.4 N

(d) If the brick falls on a foam which is 6.0 cm (thickness), we have;

Distance = 6.0 cm = 0.06 m

Therefore;

[tex]\displaystyle Force, \, F = \frac{125.568 \, J}{0.06 \, m} = \mathbf{2,092.8 \, N}[/tex]

The constant force needed to bring the brick to rest is, F ≈ 2,092.8 N

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