Work/force/kinematics slingshot problem

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An average force of 8.2 Newtons stretches a slingshot 43 centimeters to launch a 0.4 kg rock from a height of 18 meters. The spring/work equation, .5k(x)² = W, applies, with the stretch distance representing 'x'. The potential energy from the slingshot converts to kinetic energy to determine the rock's velocity just before impact. The final velocity is not zero; it is calculated using the initial velocity from the slingshot and gravitational acceleration. Time to hit the water can be found using kinematic equations.
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1. An average force of 8.2 Newtons is used to pull a .4 kg rock, stretching a sling shot 43 centimeters. The rock is shot downward from a bridge 18 meters above a stream. What will be the velocity of the rock just before it hits the water? How much time will it take to hit the water.



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3. I just want to know if the spring/work equation, .5k(x)squared=w applies to the slingshot, and would stretching it by 43 CM = the x part of the equation? I don't know what to do once I get work. How do i get the velocity of the rock? I know A after shot would be 9.8, d is 18, but is final Velocity 0?
 
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Shadowsol said:
3. I just want to know if the spring/work equation, .5k(x)squared=w applies to the slingshot, and would stretching it by 43 CM = the x part of the equation? I don't know what to do once I get work. How do i get the velocity of the rock? I know A after shot would be 9.8, d is 18, but is final Velocity 0?


It does apply, and yes. The spring PE is converted to KE. That's how you get the velocity. Final velocity nowhere is zero.
 
Ok, so I put in the force*the distance it was pulled back to get the potential sling shot energy. I than set that equal to the KE equation and got V. Than I simply used that V as V0, used 9.8 as A, and 18 m as D, than solved for V1. Is this correct?
 
Absolutely. You find the time using the same kinematics.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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