Question

0.94 Question 1 pts Trebuchets are widely recognized to be the superior siege engine. While launching its payload, the trebuchet moves a 90 kg stone a total of 25.0 m. The stone leaves the trebuchet 15.0 m above the ground where it was initially at rest. If the trebuchet can exert an average force of 2500 N on the stone during launch, how fast is the 90 kg stone moving as it is released in m/s? (Consider the force to always be in the direction of the stone's motion such that W=Fd) 312 331 27.6 20.9 37.3 413 Hide Stop sharing nard Il app.honorock.com is sharing your screen. MacBook Pro C G Search or type URL

Answer 1

The forces acting on the stone will be the force applied by the trebuchet 2500N and the force of gravity mg.

Now the applied is along the motion of the stone and travels 25m before it releases thus the work done by the force on the stone will be

W_f = 2500N x 25m = 62500 J

Now the force of gravity is against the motion of stone in vertical direction as it is going upward by 15m before it is released.So the work done by it can be given as-

W_g = -mg x 15m = -90kg x 9.8m/s² x 15m = -13230 J

So the net work done on the stone will be W_net = W_f + W_g = 62500 - 13230 J = 49270J

Now by using work energy theorem assuming the velocity when the stone is released to be v, we have

mv²/2 = W_net

=> 90kg x v²/2 = 49270 J

=> v = 33.1 m/s

So second option will be correct.