A 0.5-kg model helicopter initially at rest on the ground,starts to move directly up under the effect of a lift force of magnitude of 7.4 N.After 12 seconds the engine stops and the helicopter starts free fall.Ignoring air resistance,calculate the helicopter's velocity just as the engine stops.
A)59.9 m/s
B)178 m/s
C)48.4 m/s
D)38.0 m/s
E)118 m/s

Question

Quizplus · Accepted Answer

The helicopter accelerates upward under the influence of the lift force. The acceleration is given by:$$a = F/m = 7.4 N / 0.5 kg = 14.8 m/s^2$$The velocity after 12 seconds is:$$v = u + at = 0 + 14.8 m/s^2 * 12 s = 178 m/s$$

A 05-Kg Model Helicopter Initially at Rest on the Ground,starts to to Move