Question

Calculate the work done by a person in lifting a load of

20 kg from the ground and placing it one metre high on

a table? (g=9.8m/s2

)​

Answer 1

Explanation:

mass of the load(m)=20kg, displacement(or)height (s)=1m. Work done(W)= Force(F)*Displacement(s) force=ma=mg (acceleration due to gravity) W=Fs=mgh. =20*9.8*1. =196Nm (or) 196J.

Answer 2

Question :

Calculate the work done by a person in lifting a load of

20 kg from the ground and placing it one metre high on

a table?

Concept :

The work done is change in its energy.

Formula :

Potential Energy = mgh

Where ,

m : mass of body

g : acceleration due to gravity

h : height

Solution :

Initially the load is on ground .

$\tt \therefore$ Initial potential Energy ( $\tt {P.E.}_{i}$ ) = 0 J

After person place load one metre high on a table

Final Potential Energy ( $\tt {P.E.}_{f}$ ) = mgh

$\tt \implies {P.E}_{f}$ = 20 × 9.8 × 1

$\tt \implies {P.E}_{f}$ = 196 J

We know that

Work done = $\tt {P.E}_{f} - {P.E.}_{i}$

$\tt \implies$ Work done = 196 - 0

$\tt \implies$ Work done = 196 J

Work done = 196 J