Question

A helicopter designed for vertical takeoff has a mass of 9.2 ✕ 103 kg. Find the net work done (in joules) by all forces on the helicopter as it accelerates upward at 10.0 m/s2 through a distance of 20.7 m after starting from rest?

Answer 1

The net work done by all forces on the helicopter as it accelerates upward at
`\(10.0 \, \text{m/s}^2\)` through a distance of
`\(20.7 \, \text{m}\)` after starting from rest is approximately
`\(1.9 * 10^6 \, \text{J}\).`

Step-by-step explanation:

The work
`(\(W\))` done on an object is given by the equation
`\(W = F \cdot d\),` where
`\(F\)` is the force applied and
`\(d\)` is the distance over which the force is applied. In the case of vertical motion, the work done against gravity is expressed as
`\(W = m \cdot g \cdot h\)` , where
`\(m\)` is the mass,
`\(g\)` is the gravitational acceleration
`(\(9.8 \, \text{m/s}^2\)),` and
`\(h\)` is the height.

To find the net work done as the helicopter accelerates upward, we consider both the work done against gravity and the work done to accelerate the helicopter. The latter is given by
`\(W_{\text{acceleration}} = (1)/(2) m v^2\),` where
`\(v\)` is the final velocity. Since the helicopter starts from rest, the final velocity
`(\(v\)) is \(10.0 \, \text{m/s}\)` (the acceleration times the distance traveled).

The net work done is the sum of the work against gravity and the work for acceleration:

`\[ W_{\text{net}} = W_{\text{gravity}} + W_{\text{acceleration}} \]`

Substituting the given values
`(\(m = 9.2 * 10^3 \, \text{kg}\), \(g = 9.8 \, \text{m/s}^2\), \(h = 20.7 \, \text{m}\), and \(a = 10.0 \, \text{m/s}^2\)),` we find
`\(W_{\text{net}} \approx 1.9 * 10^6 \, \text{J}\).`This represents the total work done by all forces on the helicopter during its upward acceleration.