Calculating Work in Lifting a Bucket

[Brian McKalip]

Homework Statement

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 20.0-m tall
building. The bucket is initially at rest, but is traveling at 4.0 m/s when it reaches the top of the
building. What is the minimum amount of work that the worker did in lifting the bucket?

A) 160 J B) 3.92 kJ C) 400 J D) 560 J E) 4.08 kJ

Homework Equations

F= ma
W= Fd

The Attempt at a Solution

The answer is given to me, it is e, 4.08kJ However I am unable to get that answer,. I know that I have to solve W = Fd, and that d = distance traveled = 20m, however I'm confused by the given final velocity of 4.0m/s I don't know how to incorporate this into my process. When I solve for Force, I do F = (20kg)(-9.8m/s^2) however have a feeling, I need to determine a new acceleration from the 4.0m/s velocity given, but I don't know how to do that.

 

[rude man]

When the bucket gets to the top it has kinetic energy.

 

[Timmy1221]

Did you ever figure out how to do thins problem?

 

[rude man]

Did you ever figure out how to do thins problem?

Asking whom?

 

[wwmertww]

F= m.a
W= F.d
use thıs two fırst to get the value
K=1/2 M v^2
and add thıs one
f=20x(9,8)=196N
w=196.20=3920N
k=1/2.20.(4^2)=160N
we should add thıs two value because final velocity of the partıcle is 4 m/s.
answer=160+3920N=4080N
solving by
FUAT MERT AĞARLI

 

[haruspex]

F= m.a
W= F.d
use thıs two fırst to get the value
K=1/2 M v^2
and add thıs one
f=20x(9,8)=196N

That is not really valid since we are not told that the force is constant.
Just use net change in PE + KE.

 

[wwmertww]

A man lifts a 20-kg box from the ground up to the top of a 30-m high building. The box is initially at rest, but it
travels at 4 m/s when it reaches the top of the building. How much work is done by the man in lifting the box?
A. 5840 J
B. 160 J
C. 2400 J
D. 6000 J
E. 6160 J

answer is still same, ıt is still working if you solve it my way

 

[haruspex]

A man lifts a 20-kg box from the ground up to the top of a 30-m high building. The box is initially at rest, but it
travels at 4 m/s when it reaches the top of the building. How much work is done by the man in lifting the box?
A. 5840 J
B. 160 J
C. 2400 J
D. 6000 J
E. 6160 J

answer is still same, ıt is still working if you solve it my way

Yes, of course it gives the same answer, but if your method makes unwarranted assumptions then it is not a proof.

 

[Cutter Ketch]

Yes, of course it gives the same answer, but if your method makes unwarranted assumptions then it is not a proof.

wwmertww isn't the OP. He shouldn't be posting solutions by any method!

 

[gneill]

wwmertww isn't the OP. He shouldn't be posting solutions by any method!

The original post is several years old and the only post the OP ever made; he never returned to follow up with the thread. The mentors that have looked at the issue of the complete solution attempt offered up by @wwmertww feel that it won't benefit the OP at this late date, and that the ensuing discussion of the solution offering is probably more valuable in terms of insightful information than the original question So we've decided to let the post and the followup discussion stand.