When 100 J are put into a device that puts out 40 J, the efficiency of the device is A. 40%. B. 50%. C. 60%. D. 140%.

Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of

alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is 29.89 mi/h which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, v, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

$\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx 87.084 \ ft./s$

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s

≈ 87.084 ft./s

$v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h$

The speed of the person at the end of the impact, v₂ ≈ 59.38 mi/h

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

$v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}$

$v = \dfrac{40,000 \times 30 + 150 \times 0}{40,000 + 150} \approx 29.89$

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, the findings does not agree with the expectation

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