Find the speed with which the box hits the ground.

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Homework Help Overview

The problem involves a box with a mass of 4 kg that is lifted to a height of 2 m and then released, starting with an upward velocity of 5 m/s. The task is to find the speed at which the box hits the ground without using Newton's laws of motion.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of work-energy principles, questioning the use of initial and final velocities in the energy equations. There is confusion regarding the assumption of zero velocity at certain points in the problem.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions made regarding initial and final velocities. Some guidance has been offered regarding the application of energy conservation principles, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There is a specific instruction not to use Newton's laws of motion, which may be influencing the participants' reasoning and approach to the problem.

foreverlost
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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.
 

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