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Work

Question

A helicopter lifts a 60kg astronaut 15m vertically from the ocean by means of a cable. The acceleration of astronaut is $\frac{g}{10} m s^{- 2}$ . What is the speed of astronaut just before he reaches the helicopter $(g = 9.8 m s^{- 2})$

Moderate

A

$3.42 m s^{- 1}$
B

$4.42 m s^{- 1}$
C

$5.42 m s^{- 1}$
D

$6.42 m s^{- 1}$

Solution

$\begin{array}{l} W = ΔK . E \\ m a h = \frac{1}{2} m (v^{2} - 0^{2}) \\ \frac{9.8}{10} \times 15 = \frac{1}{2} v^{2} \\ v = \sqrt{3 \times 9.8} = \sqrt{29.4} \\ v = 5.42 {ms}^{- 1} \end{array}$

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