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Question: Worker lifts 20 kilogram bucket of concrete from ground up to top of 20meter tall building. Bucket i...

Worker lifts 20 kilogram bucket of concrete from ground up to top of 20meter tall building. Bucket initially is at rest, but travelling at 4 when it reaches top. What is the minimum amount of work that was done?
A. 3.92KJ
B. 400KJ
C. 560 KJ
D. 4.08KJ

Explanation

Solution

We will solve this question using a theorem called the work-energy theorm.1st we will define what work-energy theorem is and why we are using that theorem only. And then we will substitute all the given values to find the minimum amount of work done.

Complete answer:
Let us first look at what work-energy theorem is:
Work-energy theorem states that work done by all forces acting on a particle is equal to change in the particle’s energy.
We will see why we applied this theorem; in this question kinetic energy( KE) and potential energy(PE) both energies have been included. Kinetic energy when the body is in motion and potential energy when the body is at a particular height. Hence we will apply the work-energy theorem.
W=ΔEW = \Delta E
W=EfEi\Rightarrow W = {E_f} - {E_i}
W=(KEf+PEf)(KEiPEi)\Rightarrow W = \left( {K{E_f} + P{E_f}} \right) - \left( {K{E_i} - P{E_i}} \right)
Where KEfK{E_f}=final kinetic energy
PEfP{E_f}=final potential energy
KEiK{E_i}=kinetic energy initial
PEiP{E_i}=potential energy initial
W=12mv2+mgh0\Rightarrow W = \dfrac{1}{2}m{v^2} + mgh - 0
W=12mv2+mgh\Rightarrow W = \dfrac{1}{2}m{v^2} + mgh
Let us now substitute the values:
W=12(20)(4)2+(20)(9.8)(20)\Rightarrow W = \dfrac{1}{2}\left( {20} \right){\left( 4 \right)^2} + \left( {20} \right)\left( {9.8} \right)\left( {20} \right)
W=(10)(16)+(20)(9.8)(20)\Rightarrow W = \left( {10} \right)\left( {16} \right) + \left( {20} \right)\left( {9.8} \right)\left( {20} \right)
W=160+3920\Rightarrow W = 160 + 3920
W=4080J\Rightarrow W = 4080J
W=4.080kJ\Rightarrow W = 4.080kJ
Hence the minimum amount of work done by the worker to lift a 20 kilogram bucket of concrete from ground up to top of 20meter tall building when the bucket is initially at rest, but travelling at 4 when it reaches top is 4.080kJ4.080kJ.
Hence the correct answer to this question is option D.

Note: Note that in this question kinetic energy and potential energy both are involved, do not just consider potential energy, the answer would be different and incorrect.
We have taken the PEfP{E_f}and KEfK{E_f}as zero in this question because the mechanical energy of the bucket at the bottom is zero.
Also remember to convert the work done in joule to kilojoule by dividing the answer in joules by 1000.