A 6.1 kg flowerpot drops off the top of a tall wall. 1) 2) 3) 4) What is the mag

Question

A 6.1 kg flowerpot drops off the top of a tall wall.

1)

2)

3)

4)

What is the magnitude of the force acting on the pot while it is in the air 1.6 s after it begins to fall? After the pot has fallen 35 m, what is the magnitude of its velocity? After the pot has fallen h= 35 m, it enters a pool of 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? What is the force exerted on the liquid by the pot?

Explanation / Answer

1)force F=ma

    First, get the time it took to reach 35.0 m:
d=(1/2)at^2
t=sqrt(2d/a)=sqrt(70/9.8)=2.67
then, get the velocity of the pot at that point:
2)v=at=9.8*2.67=26.19m/s
then, assuming constant deceleration on the viscous, average velocity of the pot in the fluid is the half the sum of its initial and final velocity
v=(26.19+0)/2 = 13.09m/s
take the time it took to travel 1.5 m
t=1.5/13.09=0.114s
then get the deceleration by the formula: d=0.5at^2 --> a=2d/t^2 = 230.8m/s^2

3)deceleration=230.8 m/s^2,

then

4), the force exerted by the viscous fluid is: m*a =6.1*230.8

F=1408.1N =1400N(approx)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---