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Loading and firing a crossbow
A bowman with a crossbow shoots a 250 g bolt into the air at an angle of 30 degrees. In arming the crossbow, the bowman put the crossbow vertically downwards and exerted a force of 350 N to draw the bow back 10 cm. What is the velocity of the bolt as it leaves the crossbow if the coefficient of kinetic friction between the bolt and the 10 cm long channel that guides the bolt while being launch is mu_k=.23?
How can I get the initial energy of the bow if I don't know the spring constant? And once I find that, do I just put all that energy minus the frictional work of .23*.25 kg*g*.1 m into kinetic energy?
Customer support service by UserEcho
To answer your first question, you need to set F = k*s (F=force,k=spring constant,s=stretch from equilibrium) and solve for the spring constant, if you are looking for that initially.