In 1947, a deceleration sled was built to test the effects of extreme forces on humans and equipment. In this

sled, a test pilot undergoes a sudden negative acceleration of about 50.0 times free-fall acceleration (g). In 0.181
s, through a distance of 8.05 m, the pilot’s speed decreases from 88.9 m/s to 0 m/s. If the pilot’s mass is 70.0 kg,
how much work is done against the pilot’s body during the deceleration?

1 answer

To determine the amount of work done against the pilot's body during the deceleration, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

  1. Calculate the initial and final kinetic energies:

    • The initial velocity (\(v_i\)) is 88.9 m/s, and the final velocity (\(v_f\)) is 0 m/s.
    • The mass of the pilot (\(m\)) is 70.0 kg.

    The kinetic energy (\(KE\)) is given by the formula: \[ KE = \frac{1}{2}mv^2 \]

    • Initial kinetic energy: \[ KE_i = \frac{1}{2} \times 70.0 , \text{kg} \times (88.9 , \text{m/s})^2 \] \[ KE_i = \frac{1}{2} \times 70.0 \times 7892.21 \approx 275,000 , \text{J} \]

    • Final kinetic energy (at \(v_f = 0\)): \[ KE_f = \frac{1}{2} \times 70.0 \times (0 , \text{m/s})^2 = 0 , \text{J} \]

  2. Calculate the change in kinetic energy: \[ \Delta KE = KE_f - KE_i = 0 , \text{J} - 275,000 , \text{J} = -275,000 , \text{J} \]

    This change in kinetic energy represents the work done against the pilot's body. Since the work is negative, it means that energy is being dissipated (or done against the pilot's body).

  3. Conclusion: The amount of work done against the pilot's body during the deceleration is: \[ W = -275,000 , \text{J} \] The work done against the pilot's body is \(275,000 , \text{J}\) (magnitude), indicating the significant force experienced due to deceleration.