Find the speed with which the box hits the ground.

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The discussion revolves around calculating the speed of a box hitting the ground after being lifted and released. The box, with a mass of 4 kg and an initial upward velocity of 5 m/s, is subjected to a force of 65 N. The incorrect calculation for the final speed was found to be 6.32 m/s, while the correct speed is 8.09 m/s. The confusion stems from misapplying the concept of initial and final velocities, particularly assuming one of them must be zero. The correct approach emphasizes that the box retains a non-zero velocity when it impacts the ground.
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Homework Statement


A box whose mass is 4kg is at rest on the floor. It is then lifted to a height of 2m by a constant a constant force F and then released. At this point the box has upward velocity of 5m/s. Do not use Newton's laws of motion to solve this problem.

A. Find magnitude of the force F
F=65N

B. Find the speed with which the box hits the ground.
?

Homework Equations


g = 10m/s^2


The Attempt at a Solution



A. Wother = ΔKE + ΔPE = 65N magnitude of force

B. 0 = 1/2mvf^2-1/2m(0)^2 + mg(0) - mghi
= 1/2mvf^2 - mghi
vf = √[(2*mghi)/m] = √2*4*10*2/4 = 6.32 m/s (WRONG)

The speed is supposed to be = 8.09 m/s
I just killed the constant force in the work and substituted in the values as if the box fell to ground. I don't know what I'm missing.

I will return soon.
 
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foreverlost said:
B. 0 = 1/2mvf^2-1/2m(0)^2 + mg(0) - mghi

Why zero here?
 
Last edited:
TSny said:
Why zero here?

I assumed it had to be zero for either final or initial velocity. so i can solve for v. i first tried vf = 0m/s to say the object has landed, and then i'd solve for vi with this formula.
 
foreverlost said:
1. The problem statement, all variables and given/known

The Attempt at a Solution



A. Wother = ΔKE + ΔPE = 65N magnitude of force

B. 0 = 1/2mvf^2-1/2m(0)^2 + mg(0) - mghi
= 1/2mvf^2 - mghi
vf = √[(2*mghi)/m] = √2*4*10*2/4 = 6.32 m/s (WRONG)

The speed is supposed to be = 8.09 m/s
I just killed the constant force in the work and substituted in the values as if the box fell to ground. I don't know what I'm missing.

I will return soon.

When you solve for v...
0=½mv^2-mgh_i
½mv^2=mgh_i |:m |*2
v^2=2gh_i
v=√(2gh_i)
 
foreverlost said:
I assumed it had to be zero for either final or initial velocity. so i can solve for v. i first tried vf = 0m/s to say the object has landed, and then i'd solve for vi with this formula.

Why does one velocity have to be zero? You were given an initial velocity (vi=+5m/s), and you are told to find a final velocity (vf). By your logic, vf must be zero. This makes no sense for two reasons:
1. The ball is accelerating downwards, so as long as it starts from any height above the ground, it should have nonzero velocity when it hits the ground.
2. If you are told to find something, it is very rarely zero.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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