A 500 kg sack of coal falls vertically onto a 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no exte
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Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:
(This is correct because the horizontal motion has acceleration zero). Then:
Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:
Then, plugging in the given values, we obtain:
Finally, the effective spring constant of the firing mechanism is 1808N/m.