Calculating Force to Submerge 5 kg Object in Water

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To calculate the force required to submerge a 5-kg object that displaces 10 kg of water, the buoyant force is determined to be 98.1 N. To submerge the object, a force greater than the buoyant force must be applied. The total force needed is the buoyant force minus the weight of the object, which is 49 N. This is derived from considering both the buoyant force and the weight of the object in a free body diagram. Understanding these forces clarifies why the answer is 49 N.
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Homework Statement



A 5-kg object displaces 10 kg of water. What is the force required to submerge the object?

a. 24 N

b. 147 N

c. 98 N

*d. 49 N

Homework Equations



Buoyant Force = Weight of the Water displaced

The Attempt at a Solution



Weight of the Water displaced=10kg*9.81m/s^2=98.1N

If the buoyant force is 98.1N an equal or greater force must be applied to submerge the object.

The answer key says its 49N...why?
 
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You are forgetting about the weight of the object. If you draw a free body diagram and apply all the forces, it becomes obvious.
 
Ohh right the object pushed downward too
 
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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