Ft = m*a = 3 kg * (9.81*26/1.5) = 441.13 N

[Naeem]

Falling Flowerpot

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A 3 kg flowerpot drops from a tall building. The initial speed of the pot is zero, and you may neglect air resistance.


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a) What is the magnitude of the force acting on the pot while it is in the air 1.5 s after it begins to fall?
|F| = N *


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b) After the pot has fallen 26 m, what is its speed?
|v| = m/s *
sqrt(2*9.81*26) OK

HELP: Apply the appropriate formula from one-dimensional kinematics with constant acceleration.


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c) After the pot has fallen 26 m, it enters a viscous liquid, which brings it to rest over a distance of 1.5 m. Assuming constant deceleration over this distance, what is the magnitude of this deceleration?
|a| = m/s2 *
9.81*26/1.5 OK


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d) What is the force exerted on the liquid by the pot?
Ft = N

Need help with part d. Some guidance requested please!

 

[Galileo]

A constant decceleration means a constant force is applied to it. Use F=ma to find the force the liquid exerts on the pot.

 

[thepatientmental]

To determine the force exerted on the liquid by the pot, we can use the formula Ft = m*a, where Ft is the force, m is the mass of the pot, and a is the acceleration. Since the pot is decelerating, the acceleration will be negative. We can use the value of the deceleration calculated in part c and the mass of the pot (3 kg) to calculate the force exerted on the liquid.

Ft = 3 kg * (-6.54 m/s^2) = -19.62 N

Therefore, the force exerted on the liquid by the pot is 19.62 N in the opposite direction of the pot's motion. This force is caused by the pot's deceleration and its interaction with the liquid.