Kinetic Energy of 2kg Sinker Thrown from Bridge

AI Thread Summary
A 2kg sinker is thrown from a bridge with an initial vertical velocity of 10 m/s into a river 12m below, leading to a calculated kinetic energy of 340 J upon impact. The relevant equation for kinetic energy is KE = 1/2mv^2. Participants in the discussion highlight the need to consider both kinetic and potential energy in the overall analysis of the sinker's motion. The initial kinetic energy is derived from its velocity, while potential energy is influenced by its height above the water. Understanding both energy types is crucial for accurately determining the sinker's total energy at the moment it hits the river.
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Homework Statement


A 2kg sinker is thrown with a vertical velocity of 10m/s from a bridge into a river 12m below. What is the kinetic energy of the sinker when it hits the river??


Answer is 3.4*10^2 J


Homework Equations



KE = 1/2mv^2

The Attempt at a Solution


tried substituting directly into KE = 1/2mv^2 but obviously not write ant help would be appreciated.
 
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pat666 said:

Homework Statement


A 2kg sinker is thrown with a vertical velocity of 10m/s from a bridge into a river 12m below. What is the kinetic energy of the sinker when it hits the river??


Answer is 3.4*10^2 J


Homework Equations



KE = 1/2mv^2

The Attempt at a Solution


tried substituting directly into KE = 1/2mv^2 but obviously not write ant help would be appreciated.
Okay, so when the 'sinker' is thrown you've worked out that it has kinetic energy, but is that the only type of energy it has?
 
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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