Question

a test pilot with a mass of 70.0 kg undergoes a sudden negative acceleration of 4.90 10 2 m/s 2 . This deceleration occurs over a distance of 8.05 m. How much work is done

Answer 1

To calculate the work done on the test pilot during deceleration, you can utilize the following formula for work:

\[W = F \cdot d\]

Where:
- \[W\] is the work done (in joules, J)
- \[F\] is the force applied (in newtons, N)
- \[d\] is the distance over which the force is applied (in meters, m)

First, calculate the force (negative because it's a deceleration) utilizing Newton's second law, which states that \[F = m \cdot a\], where \[m\] is the mass and \[a\] is the expedition:

\[F = m \cdot a\]
\[F = 70.0 \, \text{kg} \cdot (-4.90 \times 10^2 \, \text{m/s}^2)\]
\[F = -34,300 \, \text{N}\]

Now that you have the force and the distance, you can calculate the work done:

\[W = (-34,300 \, \text{N}) \cdot (8.05 \, \text{m})\]
\[W = -276,315 \, \text{J}\]

So, the work done on the test pilot during deceleration is approximately \(-276,315\) joules. The negative sign denotes that work is done against the direction of kineticism, which is expected during deceleration.